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AN INTRODUCTION 



TO 



Mechanical Drawing 



BY 



FREDERIC R. HONEY, 



Instri-'ctor in Descriptive Geometry and Mechanical Drawing in 
THE Sheffield Scientific School, 

AND 

Instructor in Perspective in the Yale School of the Fine Arts. 



:fi?.ioe 50 c:e33^ts. 



NEW HAVEN, CONN.: 

The Stafford Printing Co., 393 State Street. 



Entered according to Act of Congress, in the j-ear 1881, by 

FREDERIC R. HONEY, 

in the Office of the Librarian of Congress, at Washington. 



AN INTRODUCTION 



TO 



Mechanical Drawing 



BY 



FREDERIC R. HONEY, 



Instructor in Descriptive Geometry and Mechanical Drawing in 

THE Sheffield Scientific School, 

and 

Instructor in Perspective in the Yale School of the Fine Arts, 



1 



j.k.Ul^'- 



NEW HAVEN, CONN.: 
The Stafford Printing Co., 393 State Street 

/ P ' 



Note. — The following pages are a reprint of a series of 
articles which has appeared in the Educator. The aim of 
the writer has been to simplify the subject as far as possible, in 
order to place it within the reach of those who cannot avail 
themselves of the services of a teacher. This will account for 
the minuteness of detail which may be observed throughout the 
lessons. The plates are those w^hich were designed for the 
Educator, their form and dimensions being adapted to the 
columns of that paper. F. R. H. 



'"5., 
f..'^ 



K'i 



INTRODUCTORY. 

A knowledge of Geometrical Drawing enables the student 
to represent accurately the true form and dimensions of an ob- 
ject, as well as its appearance as seen from any point of view. 
We may therefore divide the subject into two parts, viz : Pro- 
jection Drawing, and Linear Perspective. The characteristic 
difference between Projection Drawing and Perspective is this : 
by the former we supply the data, by means of which an object 
may be constructed ; while by the latter we simply represent 
the appearance of the object. An acquaintance with Projec- 
tion Drawing is therefore essential to the Engineer, the Archi- 
tect, and the Artisan. It is also desirable that the Artist 
should understand the principles of Projection, as a basis for 
the study of Perspective. Before he can accurately delineate 
an object as it appears, he must learn to represent its true 
dimensions. From the foregoing remarks, it will be seen that 
a course of instruction in Geometrical Drawing should be 
commenced by a series of lessons on Projections. We will 
first describe the instruments with which the student should be 
provided, and offer some remarks upon their use. 



CHAPTER I. 

THE INSTRUMENTS. 

The l>rawin^ Board. — a b c ^— Fig. I. — should be 
twenty -four inches long by eighteen inches wide, made of well- 
seasoned pine or other soft wood, about five-eighths of an inch 
thick. One side should be battened to prevent it from warp- 
ing, and the four edges should be straight. It is not necessary 
that the edges should form an absolutely perfect rectangle, 



since the stock of the square should be applied to only one 
edge in producing any given drawing. The surface should be 
plane, and free from holes. At a later stage in his work the 
student may, in addition to the above, provide himself with a 
board large enough for making a drawing covering twice the 
area, viz : thirty-six inches by twenty -four inches, but for the 
present the smaller one will be the most convenient size. 

The T Square.— /.—Fig. i.— should be made of well 
seasoned pear or mahogany or other hard wood, thirty-six 
inches long. It is composed of two parts, viz : the stock t and 
the blade/! The best way to construct this instrument is to 
place the blade upon the stock, to which it should be fastened 
by screws, rather than to let it into the stock. The object of 
this arrangement will be apparent on reference to the illus- 
tration, in which it will be seen that the upper surface of the 
stock becomes a continuation of the upper surface of the draw- 
ing board, and no obstruction is offered to the triangle^ which 
may slide over it. A T square constructed in this way may 
be very readily taken apart, and the edges of the stock and 
blade straightened whenever necessary. It has been already 
remarked that the stock should be applied to only one edge of 
the drawing board. It should also be noted that only one 
edge of the blade, viz : hi may be used, and in order that it 
may be apparent which edge is intended for use, the blade is 
tapered. If the edges appeared parallel, the student might be 
tempted to use either, and do inaccurate work. It is not 
necessary that the working edge of the blade hi should be 
exactly at right angles to the edge of the stock. The essential 
conditions are that the edge of the stock which is applied to 
drawing board, and the working edge of the blade should be 
straight. 

The Paper, in commencing, may be the cheaper kind. 
That known as " G. White," which may be cut into pieces 
twenty two inches by fifteen inches is a convenient size. It is, 
desirable that the length and breadth of the paper should be 
nearly equal to the corresponding dimensions of the drawing 
board, in order that the holes made by the pins, which fasten the 



5 

paper to the board may, as far as possible, be confined to one 
part of its surface. If a small piece of paper be attached to a 
large board, it is impossible to avoid making pin holes, into 
any one of which the point of the divider leg would sink, if it 
should happen that this point must be used as the centre of a 
circle. 

The Fins or Tacks should be about one-half inch 
diameter. The heads should be thin and beveled, in order 
that they may offer the least possible obstruction to the 
passage of the blade of the T square across the paper. 

The Pencil should be two or three H, or the equiva- 
lent VH, and should be cut to a fine wedge-shaped point. A 
pencil cut in this way will retain its: point longer and save the 
draughtsman a great deal of time. The lines should be drawn 
lightly in order that they may be easily erased. 

The Triangles may be made of well seasoned hard wood, 
or of rubber. The 45° triangle g — Fig. I — is a right angled 
triangle, two of its angles being equal to each other, viz : 45°. 
The length of the longest side should be from nine to twelve 
inches. The 60° triangle it — Fig. i — is a right angled triangle, 
one of its angles being 60° and the other 30°. The length of 
the longest side should be from twelve to fifteen inches. 

The Curved Kuler.— Fig II. — may be made of well 
seasoned hard wood, or of rubber, from twelve to fifteen inches 
long. It should combine short curves at the ends, with long 
curves at the middle part. 

The Ellipsograph. — Fig. III. — an instrument to be 
used in drawing ellipses, is provided with two needles which 
may be set at any required distance apart. The needle a is 
fixed to the staff/, and the needle b which is carried by the 
saddle c, is adjustable, being held to its place by the set-screw 
d. The slot in the staff/ allows the set-screw d to be moved to 
any desired position. The needles are adjusted so that the 
points a and b cover the foci of the ellipse, and then the thread 
^is extended by the pencil point to the required distance, and 
the curve described as illustrated. When one-half of the curve 
is drawn, the instrument is reversed and the other half con- 
structed. 



6 

The Scale should be made ot boxwood, twelve inches 
long, with a beveled edge divided into inches and parts. The 
beveled edge makes the use of the scale convenient and accu- 
rate, bringing the divisions close to the surface of the paper. 

The Parallel Ruler. — Fi^IV. — is a contrivance origi- 
nally designed to be used in drawing section lines, but it may 
be applied to a great variety of work, as will be evident to any 
draughtsman. The part a is an ordinary 45° triangle, having 
mounted upon it two parallel guides b and b, between which 
slides a tongue c, tapered at one end. The part/ is a straight 
edge, on which are mounted two stops d and d\ which are par- 
allel respectively to the sides of the tapered end of the tongue 
c, when the triangle and straight edge are applied to each other, 
as shown in the cut. The triangle a and the straight edge/ 
are attached to each other by the spiral spring e, the effect of 
which is to keep the stop d in contact with the corresponding 
side of the tapered end of the tongue c. In order to draw a 
series of parallel lines at equal distances apart, place the 
instrument in the position shown in the cut, and rule a line 
along the edge of the triangle gh. Hold the straight edge 
firmly in its place and slide the triangle along it, until the side 
of the tapered end of the tongue comes in contact with the 
corresponding stop d , and then allow the straight edge to be 
drawn by the spring e : the stop d is thus brought in contact 
with the corresponding edge of the tongue. Rule the second 
line along the edge gh, which will evidently be parallel to the 
first. The operation may be repeated, and any number of 
parallel lines drawn, and the distance between them will be 
regulated by moving the tongue as far as may be desired into 
the space between the stops d and d' . 

The India Ink may be either liquid or in cake. The 
former is recommended, since it is prepared expressly for the 
purpose. Before using it, it is desirable to shake the bottle, 
in order that the heavier and lighter parts may be mixed 
together. From time to time one or two drops of water should 
be added to supply the loss by evaporation. If the cake be 
employed a saucer will be needed, to which a few^ drops of 



water should be applied. The cake should be gently rubbed 
on the surface, which must be smooth, and a little water added 
from time to time, until the requisite amount has been pre- 
pared and the ink is black enough for use. 

The India Rubber should be soft, in order that in 
erasing a pencil line the surface of the paper may be raised as 
little as possible. 

The DiTiders should be provided with a lengthening 
bar, a pencil holder and a pen. The needle point is recom- 
mended. 

The Ruling* Pen should have one nib on a hinge, in 
order that it may be easily cleaned after being used. Before 
supplying the pen with ink, bring the nibs close together by 
means of the set-screw, and with an ordinary writing pen 
place between the nibs the necessary quantity of India ink, 
The charge should be large, in order that the flow of ink from 
the pen may be steady, and should be renewed before it is 
exhausted. Care should be taken to confine the ink to the 
space between the nibs, in order to avoid the possibility of its 
being transferred to the straight edge or curved ruler in 
drawing a line. 

The instruments which have been described are all that are 
necessary for the present. When the student has made some 
progress he may provide himself with a small bow pencil and 
pen, the spring instruments, a pair of hair dividers, and beam 
compasses. 

We will conclude this chapter with some practical remarks 
on the adjustment of the instruments. 

In attaching the papery/^/;/^ — Fig. I — to the board, it is nec- 
essary that it shall lie smoothly and be firmly secured, in order 
that after the drawing is commenced there may be no possi- 
bility of its shifting. Put one pin / in its place, stretch the 
paper diagonally from this corner and secure the opposite 
corner /, being careful at the same time to apply the stock of 
the T square to the left side of the board, and see that the 
edge of the paper jk coincides with the working edge of 
the blade of the T square. Secure the remaining corners k 
and m, being careful to stretch the paper, and bring the heads 



8 

of the tacks down upon it. It should be remembered that the 
stock of the T sqiiare should be applied only to the edge of the 
drawing board on the left hand of the student, as shown in the 
cut. Lines parallel to jk should be ruled along the edge hi in 
any part of the paper, the stock e being held firmly against the 
board, and lines perpendicular to these should be ruled along 
a perpendicular edge of either of the triangles, as ;/, the triangle 
being applied to the blade of the T square, as shown in the 
illustration. From this it will be clear that it is not neces- 
sary that the edges of the board should form a perfect rect- 
angle, nor that the working edge of the blade of the T square 
should be perpendicular to the working edge of the stock. 
Attention is specially called to these points, since draughts- 
men often spend time unnecessarily in correcting errors of 
this kind. But since we may decide to use either edge of the 
board, it is important that they, as well as the working edges 
of the stock and blade of the T square, shall be straight.. In 
order to ascertain if the working edge of the blade of the T 
square be straight, rule a fine line o p q along it, then turn it 
over, that is, bring the upper surface of the blade into coinci- 
dence with the paper, passing the edge o r q through the 
points ^ and ^. \i o r q does not coincide with ^/ ^, it will 
be apparent that the edge of the blade should be straightened. 
In order to ascertain if the edges of the triangle which are 
supposed to contain the right angle are perpendicular to each 
other, apply the triangle to the T square, as shown at g, then 
turn it over into the position g\ and if the lines ruled along 
the edge which is supposed to be perpendicular to the edge 
of the blade of the T square do not coincide, as shown in the 
illustration, the edges of the triangle are not perpendicular to 
each other, and the error should be corrected before the in- 
strument is used. After applying this test to the 45° triangle 
and correcting the error, if any, ascertain if the remaining 
angles are equal, as follows : Place the triangle in the position 
t s 21, and rule a line st ; then place the triangle in the position 
/ f u', and if s' t' does not coincide with / s, it will be appar- 
ent that the angles are unequal, and the error should be 



corrected. After applying the first test to the 60° triangle, and 
correcting the error, if any, ascertain if the remaining angles are 
respectively 60° and 30°, as follows : Place the triangle in the 
position 71, and rule a line vw ; then place the triangle in the 
position n\ and rule a line w' v' ; from v, the point of inter- 
section of these lines, draw a line v x along the edge of the 
T square ; with •z/ as a centre, and with any convenient radius 
describe an arc, cutting the lines v w, w' v' and v x respect- 
ively in the points w, v' and x. If the angles are correct the 
point v' will' bisect the arc w x. 



CHAPTER 11. 

PROJECTIONS. 

It was stated in the introduction that a knowledge of 
Geometrical Drawing enables the student to represent accu- 
rately the true form and dimensions of an object. This is 
accomplished by means of its projections, or in ordinary 
language, by its plan and elevation. In order to show all the 
dimensions of an object, we give two views of it, or its repre- 
sentation upon two planes. These planes are taken at right 
angles to each other. 

PLATE A. 

Fig. I. — a big represents a vertical plane, and is called 
the vertical plane of projection, g I c d represents a horizontal 
plane, and is called the horizontal plane of projection, g /, the 
the .line of intersection of these planes, is called the groicnd 
line. Let it be required to represent a point P in space. Let 
fall a perpendicular from P to the horizontal plane, r the foot 
of this perpendicular is the projection of the point upon the 
horizontal plane, and is called its horizojttal projection or plan. 
Also let fall a perpendicular from Pto the vertical plane, q the 
foot of this perpendicular is the projection of the point on the 
vertical plane, and is called its vertical projection or elevation. 
If the plan and elevation of a point be given the point is 
located in space-. For if we erect from q a perpendicular to 



lO 



the vertical plane, and from ra perpendicular to the horizontal 
plane, the intersection of these perpendiculars at P will 
determine the position of the point in space. We may con- 
sider the planes a b I g and g I c d to represent a sheet of 
drawing paper bent at the line^ /, one part of the paper being 
at right angles to the other. Now since it would be impracti- 
cable to use the drawing paper in this way, we conceive, after 
the projections of the point are determined as described above, 
that the vertical plane a b I g\s revolved around the ground 
line ^/ as an axis, until it reaches the position a' b' I g, when it 
becomes a continuation of the horizontal plane g I c d. In this 
position both planes may be represented upon the same plane or 
sheet of paper a' b' c d. It will be observed that during the revo- 
lution every point in the vertical plane describes a quadrant of a 
circle, q the elevation of the point describes a quandrant q s 
and the elevation falls at s. If we let fall from q a perpendicu- 
lar to the horizontal plane it will pierce it in the ground line 
at t, and if we complete the rectangle P q t rwQ see that s t= 
q i=P r; that is, the distaitce from the elevation of the poi7tt s 
to the ground line is equal to the distance fj^om the point in space 
P to to the horizontal plane. Also r t^=P q, that is, the distance 
from the plan of the point r to the ground line is equal to the 
distance from, the point in space P, to the vertical plane. 

Let it be required to represent a point at a distance of three 
inches from the vertical plane, and four inches from the horizon- 
tal plane. Let e f h i — Fig. II. — represent the drawing board 
and a' b' c d the drawing paper attached to the board, as de- 
scribed in Chap. I. It should be remembered that the edge of 
the board e i should be on the left hand of the student, as is illus- 
trated in Plate I, which is lettered in the same way as the 
paper in Fig. II, in order that they may be compared. 
Apply the stock of the T square to the edge of the board e i, 
and rule a line ^ / along the edge of the blade. The part of the 
paper a' b' Ig will represent the vertical plane a b Ig — Fig. I, 
and the part g Ic d will represent the horizontal plane g I c d — 
Fig. I. Mark s the elevation of the point, at a distance from^ / 
equal to four inches, the distance from the point to the hori^ 



II 



zontal plane. Apply one of the perpendicular edges of either 
triangle to the edge of the blade of the T square, so that the 
other perpendicular edge will pass through s, and rule a line 
s tr along it, perpendicular to the ground line. From / on this 
line lay off / r, equal to three inches, the distance from the 
point to the vertical plane, r is the plan of the point 



PLATE I. 

In this and the succeeding plates the dimensions and posi- 
tions of the figures will be given, which the student should 
draw according to the directions. The paper should be twen- 
ty-two inches long by fifteen inches wide. Commence by 
drawing the ground line £- 1 across the middle of the paper, 
and it should be remembered that the part a! b' I g represents 
the vertical plane, and the part g I c d represents the hori- 
zontal plane, and that while these two parts are in the same 
as plane, they are supposed to be at right angles to each other 
shown in Fig. I, Plate A. If this be continually borne in mind, 
the student will have little difficulty in understanding the 
constructions which follow. 

Prob. I. — To draw the plan and elevation of a straight line 
five itzches long, which is perpendicular to the horizontal plane. 
The plan of a straight line, which is perpendicular to the hor- 
izontal plane P r — Fig. I, Plate A — is a point r. The plan 
of a line is a line passing through the horizontal projections of 
all of its points, and in this particular case all the points are 
evidently projected in the same point r, that is, the plan of the 
line is a view of it, looking at it in the direction of the ver- 
tical arrow. The eye of the observer is supposed to be in a 
continuation of the line, and it is plain that only a point will be 
visible. The elevation q t=s t is equal to the line, and is a 
view of it looking at it in the direction of the horizontal arrow. 

Therefore assume a point e' at a distance of two inches and 
a half from the left side of the paper, and four inches 
from the ground line. From e' draw e' e perpendicular to the 
ground line, and lay ofi" upon it from the ground line e" e^ 



12 

equal to five inches, e and d 'e are the required plan and ele- 
vation. 

Prob. 11. — To draw the plan and elevation of a square prism, 
one of whose bases is situated in the horizontal plaiie. Length of 
the side of the base^2\" . Altitude of p7isin=^" . Assume a 
point / at a distance of seven inches from the left side of 
the paper, and of five and a half inches from the ground line. 
From / draw e' f and e' h', forming with the ground line re- 
spectively angles of 30° and 60°. Lines forming these angles 
with the ground line should be ruled along the edge of the 60° 
triangle applied to the edge of the blade of the T square, as 
indicated by the broken lines. Lines drawn in this way will 
be perpendicular to each other at the point e'. From / lay 
off^'/'and/// each equal to two inches and a half — the 
length of the side of the base. From/ draw/' i' parallel to ^ 
h', and from // draw h' i' parallel to e' f, intersecting/' i' in 
i'. These parallel lines may be ruled along the edge of the 60° 
triangle, sliding it along the edge of the blade of the T square 
into the desired position, as indicated. The square e'f i' h' 
is the plan of the prism. Since the altitude of the prism is 
five inches, each perpendicular edge is five inches long, and 
the elevation will be determined as follows : From the points 
e' ,f' , i', and h' , respectively, draw e'e, ff, i'i and h'h per- 
pendicular to the ground line, and lay off from the ground line 
e"e,f'f ii'i and JL"h, each equal to five inches. Draw a line 
from h to/ parallel to the ground line, which will evidently 
pass through the points e and i. hf is the elevation of the 
upper base of the prism, and that part of the ground line which 
lies between the points Ji" and /" is the elevation of the 
lower base, which is situated in the horizontal plane. Jiff'h" 
is the elevation required. The ground line is the elevation of 
the horizontal plane, and therefore contains the elevation of 
every point which is situated in the horizontal plane, and con- 
sequently all surfaces which coincide with it. Also the 
elevations of all horizontal planes, as the upper base of 
the prism, will be straight lines parallel to the ground line. 
The vertical edge, whose plan is /', will be invisible in the 



13 

elevation i i" , and will be represented by a broken line. This 
will be plain if we consider the direction in which the eye of 
the observer is supposed to be viewing the object. If we refer 
to Fig. I, Plate A. we see that the horizontal arrow, which in- 
dicates the direction in which the object is seen, corresponds 
with the arrow in Prob. II, and that only those vertical edges 
whose plans are h' , e^ and/' will be visible in the elevation. 

Prob. III. — To draw the plan and elevation of an hexagojial 
pyramid. Diagonal of hexagon=^}f\" . Altitude of pyramid=^ 
5''. Assume a point o' at a distance of nine inches and a half 
from the right side of the paper, and four inches from the ground 
line. With o' as a centre, and with a radius equal to one inch 
and three-quarters, one-half the length of the diagonal, describe 
2i circXe. e' f k' i' f k'. Commencing at any point ^' on the 
circumference, divide it into six equal parts, in the points e',f', 
k'y i'yf and k'. Draw the sides of the base, e' f',f' h', h' i', i' 
f,f k', and k' e', each of which will be equal to the radius o' e'. 
Also draw the diagonals e' i' J'j' , and h' k' . o' — e' f h' i' f k' 
is the plan of the pyramid. The point o' is the plan of the vertex, 
and the lines o' e', o'f, o' h' , o' i\ o' f and o' k' the plans 
of the edges connecting the vertex with the angular points of 
the base. From o' draw o' o perpendicular to the ground line. 
Lay off m o equal to five inches, the altitude of the pyramid. 
From e' ,f\ h' , i',f and k' , respectively, draw e' e, f'f h' h, 
i' ij'j and k' k, perpendicular to the ground line, intersecting 
it in ^,/, /2, /, 7 and /&, the' elevations of the angular points oi 
the base. Join o with each of these points, o e i'ls the ele- 
vation of the pyramid. The visible edges are o e, of o h, and 
i, and are represented by full lines, and the invisible edges, 
oj and k, are represented by broken lines. 

Prob. IV. — To draw the plan and elevation of an hexagonal 
prism staiiding itpon an octagonal prism. Make the diagonal of 
the hexagon 2^" , that of the octagon 4", the altitude of the 
hexagonal prism 2", and the height of the octagonal prism 3". 
Assume a point <? at a distance of four inches from the right 
side of the paper, and four inches from the ground line. 



14 

With d? as a centre, and with a radius equal to li", one half the 
diagonal of the hexagon, describe a circle and construct a 
regular hexagon e'f h' i'f k' as described for the preceding 
problem. With £? as a centre and with a radius equal to 2", 
one half the diagonal of the octagon, describe a circle. Divide 
the circumference of this circle into eight equal parts, in 
the points m\ n',p', q' , r', s', t', and v'. Six of these points 
may be determined, by the placing the 45° triangle, in the 
different positions i, 2 and 3, in contact with one of the 
straight edges of the 60" triangle w, passing the edges of the 
triangle through the centre of the circle 0, and marking the 
extremities of the diameters, m' , n', p' \ 7-', s' and /'. A line 
through 0, parallel to the straight edge w, will cut the circum- 
ference in the remaining points v' and q' . Join m' n' &c., 
&c., e* f h' i' f k' — in' n' p' q' r' s' f v', is the plan of the prisms. 
From v', in' &c., &c., draw v' z>, in' m, &c., &c., perpendicular 
to the ground line, and lay off from the ground line v" v, in" 
in &c., &c., each equal to 3'', the altitude of the octagonal 
prism. From v draw v q parallel to the ground line. From 
e' , f &c., draw e' e, f f ^c, perpendicular to the ground line, 
and from v q lay o^e" e,f"f &c., each equal to 2", the altitude 
of the hexagonal prism. From e draw e i parallel to the ground 
line, e i q q" v" v is the elevation of the prisms. In this fig- 
ure omit those lines which are the elevations of the invisible 
edges. 

After making all the constructions in pencil we may ink the 
lines. The T square may be dispensed with, and a straight 
edge of either of the triangles may be used in ruling the lines. 
The ruling pen should be set for a fine line, and should be 
tested upon a separate piece of paper before applying it to the 
drawing. Having set the nibs at the desired distance apart, 
the pen should not be altered until the drawing is complete, in 
order that all the lines may be of the same thickness. In ruling 
a line the pen should be held in a position perpendicular to the 
surface of the paper, and care should be taken to prevent the 
ink from coming in contact with the straight edge. This will 
be accomplished by placing the straight edge a very little 
distance from and parallel to the line. After all the Hnes are 



15 

drawn as indicated in the plate, the pen should be set for a 
heavy line, and the ground line^/ should be inked. 

PLATE II. 

Prob. I, — To draw the elevation and the plan of a straight 
line jive inches long, perpendicular to the vertical plane, 
ojte extremity of the line being at a distance of one ifich from the 
vertical plane. The elevation of a straight line which is 
perpendicular to the vertical plane is a point. The line Pu — 
Fig. I. — Plate A is supposed to be perpendicular to the ver- 
tical plane a b I g, and every point of the line is evidently pro- 
jected in the same point q, that is, the elevation of Pn is q. 
In drawing the elevation of an object the eye of the observer 
is supposed to be looking at it in the direction of the horizon- 
tal arrow, that is, in a direction perpendicular to the vertical 
plane. The eye of the observer is supposed to be in a contin- 
uation of a line uP, and it is plain that only a point will be 
visible. Since the line is perpendicular to the vertical plane 
it is parallel to the horizontal plane, and is projected upon 
it in its true length. Thus rv, the plan of the line, is equal to 
Pu, the line in space. P71 and r v are opposite and there- 
fore equal sides of the rectangle Pn. v r. The plan of the 
line is a view of it looking at it in the direction of the ver- 
tical arrow. 

Therefore assume a point a a.t sl distance of three inches 
from the left side of the paper, and of four inches from the 
ground line. From a draw a a" perpendicular to the ground 
line. Lay oft' from the ground line to a' a distance equal to 
one inch, the distance from one extremity of the line to the 
vertical plane. Also lay oft" a' a" equal to five inches, the 
length of the given line. The point a is the elevation, and a' 
a" is the plan of the line. 

Prob. II. — To draw the elevation and the pla7i of a triangular 
prism whose bases are parallel to the vertical plane, one base being 
at a distance of one inch from, the vertical plane. Length of sides 
of base^=^i" , 3I" and 4" respectively. Length of prism^=s"- 



x6 

Since the bases of the prism are parallel to the vertical 
plane, the edges which are perpendicular to the bases are 
perpendicular to the vertical plane, and will therefore be 
projected upon it in points as in the preceding problem. 
Since the bases of the prism are parallel to the vertical plane 
the sides of the bases are also parallel to the vertical plane, 
and will therefore be projected upon in their true length. 
This will be plain if we refer to Fig. i — Plate A — Pii. is 
parallel to the horizontal plane, and is projected upon it in its 
true length rr-. 

Draw the elevation of the prism as follows : assume a point 
^ at a distance of six inches from the left side of the 
paper and of four inches from the ground line. From a draw 
a b, three inches long, forming with the ground line an angle of 
30°. On ^ ^ as a base construct a triangle a b c, a c being 
equal to four inches, and b c equal to three inches and 
three-quarters. The point c may be found as follows : with a 
as a centre and a radius equal to four inches describe an arc, 
and with ^ as a centre and a radius equal to three inches and 
three-quarters describe an arc intersecting the former in r. a b 
c is the elevation of the prism. From a, b and c, respectively 
draw a a" , b b" and c c" , perpendicular to the ground line. Lay 
off from the ground line to a' a distance equal to one inch. 
From a' draw a' c' parallel to the ground line, intersecting bb" 
in b' and cc" in c' . The points a' , b' and c' are the plans of the 
extremities of the edges which are perpendicular to the vertical 
plane and the line a' c' is the plan of a base of the prism. 
From a' lay off a' a" equal to five inches, the length of the 
prism. From a" draw a" c' parallel to the ground line, inter- 
secting b'b" in b" , and c'c" in c" . The lines a'a" , b'b" and c'c" 
are the plans of the edges which are perpendicular to the 
vertical plane, and the line a" c" is the plan of a base ; a' a" c" 
c' is the plan of the prism. 

Prob. TIL — To draw the elevation and plan of a hollow rigkt 
circular cylinder whose bases are parallel to the vertical plane, 
one base being at a distance of one inch from the vet'tical plane. 
Diameter of cylinder^=^^2'' \ diameter of hole^=i2^" . Length of 
cylinder=-^" . 



17 

Assume a point ^ at a distance of nine inches from the right 
side of the paper, and of four inches from the ground line. 
With ^ as a centre and a radius equal to one inch and a 
quarter, one half the diameter of the hole, describe the circle 
cd. With the same centre, and a radius equal to one inch 
and three quarters, one half the diameter of the cylinder, 
describe the circle a b. 

Since the bases of the cylinder are parallel to the vertical 
plane, the axis is perpendicular to the vertical plane, and is 
projected upon it in a point. The point o is the elevation of 
the axis. The elements of the cylinder being parallel to the 
axis are perpendicular to the vertical plane, and are therefore 
projected upon it in points, and since the elements are equidis- 
tant from the axis they are projected in the circumference of a 
circle. The circle a b\?> the elevation of the outer surface of 
the cylinder, and the circle c d i?, the elevation of the inner 
surface ; the area between the circles represents the material, 
of which the thickness is equal to a c. From the extremities 
of the horizontal diameters of the circles a, c, d and b respect- 
ively, draw aa", cc" , dd" and bb" perpendicular to the ground 
line. On aa" lay off from the ground line to a' a distance 
equal to one inch, the distance from one base of the cylinder 
to the vertical plane. From a' draw a'b' parallel to the 
ground line, intersecting cc", dd" and bb" respectively in 
c' , d' and b' . a'b' is the plan of one base of the cylinder. 
From a' lay off a'a" , equal to five inches, the length of the 
cylinder, and from a" draw a" b" parallel to the ground line, 
intersecting c'c" , d'd" and b'b" respectively in c" , d" and b" . 
a" b" is the plan of one base of the cylinder, a'b' b" a" is the 
plan of the cylinder ; a' a" and b'b" are the plans of the 
extreme visible elements of the outer surface ; c'c" and d'd" 
are the extreme elements of the inner surface, and* are repre- 
sented by broken lines because they are invisible. 

Prob. IV. — To draw the plan of a solid, whose elevation is 
given ; one base being at a distance of one inch from the vertical 
plane. Lejtgth of solid=$". 



i8 

Draw the elevation as follows : assume a point a d.t a. dis- 
tance of five inches from the right side of the paper, and of 
three inches and a half from the ground line. From a draw a b, 
three inches long, forming with the ground line an angle of 30°. 
From a and b respectively draw ag and bh, forming with the 
ground line angles of 60°, that is draw ag and bh perpendicular 
to ab ; and make ag equal to one half inch. Complete the 
rectangle abhg. On ab lay off ac and bd, each equal to one 
inch and a quarter. From c draw ce, forming an angle of 60° 
with the ground line, and make ce equal to one inch and a 
quarter ; produce ec to k, making ik equal to one inch and a 
quarter. Also from d draw df, forming an angle of 60° with 
the ground line ; produce fd to / and complete the rectangle 
eflk. It will be observed that all the lines in this figure may 
be ruled along the edge of the 60° triangle applied to the T 
square, as is illustrated in Prob. II, Plate I. The figure afhk 
is the elevation which is given. The points a, c, e, &c., &c., 
are the elevations of edges which are perpendicular to the 
vertical plane. 

From a, e, /, d, b and h respectively draw aa" ^ ^/', &c., 
&c., perpendicular to the ground line. On aa" lay off" 
from the ground line to a' a distance equal to one inch. 
From a' draw a'h' parallel to the ground line, intersecting hh" 
in h'. Lay off a' a'' equal to five inches, the length of the 
solid, and from a" draw a"h" parallel to the ground line, in- 
tersecting hh" in h" . a'h'h"a" is the required plan. 

It will be observed that we have represented only those 
edges which are visible from the point of sight. The eye of 
the observer is supposed to be viewing the object in the 
direction of the arrow which corresponds with the vertical 
arrow in Fig. I, Plate A. 

EXERCISES. 

The following exercises are given as tests of the student's 
knowledge of the constructions contained in Plates I and II. 

Exercise I. — Draw the elevatio7i of a square pyramid stand- 
ing icpoiz a square pedestal. Altitude of pyra7nidz=:^" ; thickness 
of pedestal=\". Assume the plan as follows : draw a'b'y four 



19 

inches long, forming with the ground line an angle of 30°. 
On a'b' construct the square a'b'c'd! . This may be done by 
drawing a'd' and b'c' , each forming an angle of 60° with the 
ground line, and four inches long, and c' d! , forming an angle of 
30° with the ground line. Draw the lines e'f and^'/^' parallel 
to a'b' y and at distances from it respectively of \" and -^h" - 
Also dxdiwfg' and h'e' parallel to b'c', and at distances from it 
of h'' and 2i^". Draw the diagonals e'g' and fk'. a'b'c'd' is 
the plan of the pedestal, and 0' — e'f'g'h! the plan of the pyramid. 

Draw the elevation, representing the visible edges by full 
lines, and the invisible edges by broken lines, remembering that 
the direction in which the object is seen is that indicated by 
the arrow, which corresponds with the horizontal arrow in Fig. 
I, Plate A. Since the lower base of the pedestal is situated in 
the horizontal plane, its elevation will be in the ground line ; and 
since the thickness of the pedestal is one inch, the elevation 
of the upper base will be a straight line parallel to the ground 
line, one inch from it ; this line will also contain the elevation 
of the base of the pyramid, because the base of the pyramid is 
in the plane of the upper base of the pedestal, and since the 
altitude of the pyramid is five inches, the elevation of the 
vertex will be a point five inches above the elevation of the 
upper base of the pedestal, or six inches from the ground line. 

Exercise IT. — Draw the plan of a solid whose elevation is 
given. 

Assume the elevation of the solid as follows : with a centre 
describe circles of 3", 4^" and 6" diameter; within the circle 
abcdefgh inscribe the regular octagon abcdefgh ; within the 
circle ijklmn inscribe the regular hexagon ijklmn, and within 
the circle pqr inscribe the equilateral triangle /^r. The octa- 
gon is the elevation of a prism whose length is three inches, 
the hexagon is the elevation of a prism whose length is two 
inches, and the triangle is the elevation of a prism whose 
length is one inch. The bases of the prisms are parallel to 
the vertical plane ; one base of the triangular prism is in the 
plane of a base of the hexagonal prism, and one base of the 
hexagonal prism is in the plane of a base of the octagonal 



20 

prism, that is the prisms together form a solid. The edges 
which are represented by the points a, b, c, &c., i,j\ k, 8ic.,p, q, r, 
are perpendicular to the vertical plane. Draw the plan and 
represent the visible lines only, remembering that the direction 
in which the object is seen is that indicated by the arrow, which 
corresponds with the vertical arrow in Fig. I, Plate A. 

PLATE III. 

Prob. I. — To draw the plan and the elevation of a straight line 
jive inches long, which is parallel to the horizontal plane and at 
a distance of four inches from it, and which forms with the 
vertical plane an angle of 60°. Since the line is parallel to the 
horizontal plane it will be projected upon it in its true length, 
and the plan will form with the ground line an angle equal to 
that formed by the line with the vertical plane. And since 
the line is parallel to the horizontal plane all its points are 
equidistant from the horizontal plane, and will be projected on 
the vertical plane at equal distances from the ground line, that 
is, the elevation will be a straight line parallel to the ground 
line. Therefore assume a point a' at a distance of two inches 
from the left side of the paper, and of two inches from the 
ground line. From a' draw a'b' five inches long, the length 
of the given line, forming with the ground line an angle of 60°. 
Draw ab parallel to the ground line, at a distance of four inches 
from it, the distance from the line to the horizontal plane. 
From a' draw a' a, and from b' draw b'b perpendicular to the 
ground line, intersecting ab respectively in a and b, the eleva- 
tion of the extremities of the line, a'b' is the plan, and ab 
the elevation of the given line. 

Prob. II. — To draw the plan and the elevation of a square 
prism, whose bases are perpendiciilar to the horizontal plane, the 
edges which are perpendicular to the bases forming with the 
vertical plane angles of 60°. Length of side of base=2" ; length 
' of prism=z^" . First draw the elevation and plan of the prism 
when the edges are perpendicular to the vertical plane. As- 
sume a point a No. i at a distance of eight inches from the 
left side of the paper, and four inches from the ground line. 



21 

From a draw ab, two inches long, the length of the side of the 
base, forming with the ground line an angle of 30°. On ab 
construct the square abed. From a,b,c, and d, respectively, 
draw aa" , bb" , cc" and dd" perpendicular to the ground line- 
Lay off from the ground line to a! a distance equal to one 
inch and a half Draw a'c' parallel to the ground line, the 
plan of one base of the prism. Lay off a' a" equal to five 
inches, the length of the prism and draw a"c" parallel to the 
ground line, the plan of the other base ; abed is the elevation, 
and a'e'e"a" the plan of the prism. 

Now draw the plan and the elevation when the edges form 
with the vertical plane angles of 60°. Assume a point a'" No_ 
2, at a distance of seven inches from the right side of the 
paper, and two inches and a half from the ground line ; from 
a'" draw a' "a"" equal to a' a" No. i, forming with the ground 
line an angle of 60°, equal to the angle which the edges of the 
prism form with the vertical plane. From a'" draw a'"e'" 
equal to a'e\ forming an angle of 30° with the ground line, 
that is, perpendicular to a'"a"" . Also lay off a"'d'" equal to 
a'd' and a"'b"' equal to a'b' . From a"" draw a""e"" paral- 
lel to a"'e"', and from e'" , b'" and d'" , respectively draw c'" 
e"", b'"b"", and d"'d'"\ parallel to a'" a"". a'"c'"e""a"", is 
the plan of the prism which is simply a copy of a'e'c"a" . The 
edges which are perpendicular to the bases are parallel to the 
horizontal plane ; their elevations will therefore be parallel to 
the ground line, and their distances from the ground line will 
be equal to the distances respectively from a,b,e and d, to the 
ground line. Therefore from a,b,c and d, respectively draw 
aa^^, bb^f, ee^^ and dd^ parallel to the ground line. From a'" 
draw a'"a, and from a"" draw a""a^^ perpendicular to the 
ground line, intersecting a^a^^ in the points a^ and a^^. From 
b'" draw b'"b, and from b"" draw b""b,, perpendicular to the 
ground line, intersecting b^b^^ in the points b, and b,^. From 
c'" draw /'V^ and from e"" draw e""c^^ perpendicular to the 
ground line, intersecting e^e,, in the points e, and e^^. From 
d'" draw d'"d^ and from d'"' draw d''"d^^ perpendicular 
to the ground line, intersecting d^d^^ in the points d^ and d^^. 



22 

]om a^^b^^, b^^c^^,c^^d^^ 2ir\A d^^a^^. Also join afbf,bfC„c,d,2Xi^ 
dfttf. a^bfiffCf^dftdt is the elevation required. It should be 
observed that a^^b^^,a^b^,d,Cf and df,c^,zx& parallel to each other ; 
also that a^^d^f , b^^c,, , a^d^ and b^c, are parallel to each other. 
The lines b^c^, c^d, and c^Cf, are drawn broken, because they are 
invisible from the point of sight. This will be plain if we 
consider that the direction in which the object is seen is that 
indicated by the arrow. 

PLATE IV. 

Prob. I. — To draw the elevation a^td plan of a straight line 
six itiches long, which is pa^'allel to the vertical plane, and at a 
distafice of four inches from it, and which forms with the hori- 
zontal plane an angle of 60°. Since the line is parallel to the 
vertical plane it will be projected upon it in its true length, 
and the elevation will form, with the ground line, an angle 
equal to that formed by the line with the horizontal plane. 
And since the line is parallel to the vertical plane all its points 
are equidistant from the vertical pjane, and will be projected 
on the horizontal plane at equal distances from the ground 
line ; that is, the plan will be a straight line parallel to the 
ground line. Therefore assume a point a in the ground line, 
at a distance of five inches from the left side of the paper. 
From a draw ab six inches long, the given length of the line, 
forming with the ground line an angle of 60°. Draw a'b' par- 
allel to the ground line, at a distance of four inches from it, the 
distance from the line to the vertical plane. From a draw aa' , 
and from b draw bb' perpendicular to the ground line, inter- 
secting a'b' respectively in a' and b' , the plans of the extremi- 
ties of the line ; ab is the elevation and a'b' the plan of the 
given line. 

Prob. II. — To drazv the elevatiofi and the plan of an hex- 
agonal pyra^nid, whose axis is parallel to the vertical plane, and 
whose base forms with the horizontal plane an angle of 30°. 
Length of side of hexagon=2". Altitude of pyramid=6". 
First draw the plan and elevation of the pyramid when the 
•base coincides with the horizontal plane. Assume a point 



23 

o' No. I at a distance of eleven inches from the left side of 
the jDaper, and of four inches from the ground line. With a 
centre o' , and a radius equal to two inches, describe a circle 
a'b'c'd'e'f, and within it inscribe the regular hexagon a!b'c'd'e' 
f. Draw the diagonals of the hexagon, viz. a! d' , b'e' and c'f. 
o'-a'b'c'd'e'f is the plan of the pyramid. From o* draw o'o per- 
pendicular to the ground line, and lay off from the ground line 
to 0, a distance equal to six inches, the altitude of the pyramid. 
From the points a\ b' , c' , d\ e' and / draw perpendiculars to 
the ground line, intersecting it in the points a, b, c, d, e and/ 
Join oa, ob, oc, od, oe and of. oad is the elevation of the 
pyramid. 

Now draw the elevation and the plan when the base forms 
with the horizontal plane an angle of 30°. Assume a point 
a, No. 2 in the ground line, at a distance of five inches from 
the right side of the paper. From a^ draw a^d^ equal to 
ad No. I, forming an angle of 30° with the ground line. 
Make ap^^ equal to ao,. From 0^^ draw o^^o^^^ perpendicular 
to a^d^ , that is, draw o^^o^tt , forming an angle of 60" with the 
ground line. Make 0^,0,^^ equal to 0^0. Also lay oE a,f, equal 
to af; afi^ equal to ab ; a^e, equal to ae, and a^c^ equal to ac. 
Join 0^^^ with a^,f^,bj,e^,c^ and dp^^^. d^a^, which is a copy of oad, 
No. I, is the elevation of the pyramid. In No. 2 we must un- 
derstand that the pyramid is supposed to be moved into a 
* position in which the plane of the base forms with the horizon- 
tal plane an angle of 30°, each point of the solid being at a 
distance from the vertical plane, equal to the distance from the 
vertical plane of the corresponding point in No. I. That is, 
the plan of the vertex, and of each of the angular points of 
the base will be found at a distance from the ground line 
equal to the distance from the ground line respectively of 
0' ,a' ,b' ,c' ,d' ,€" and/ No. i. Also the plan of the vertex and of 
each of the angular points of the base will be found in a per- 
pendicular to the ground line drawn from Oii,,ai,b„c„dne, and 
/ No. 2. We may therefore make the following construc- 
tion : from 0' draw o'o" parallel to the ground line. From 
0,1, draw o,,,o" perpendicular to the ground line, intersecting 



24 

the former in o" the plan of the vertex. From a' ,b' ,c' 4' ,e' and 
f respectively, draw a' a" ,b' b" ,c' c" ,d' d" / e" and f'f" parallel 
to the ground line. From a„b„c„d„e,2(.ndf, draw perpendiculai-s 
to the ground line intersecting a' a" ,b' b" ,c' c" ,d' d" 4 e" and 
f'f" respectively, in a" ,b" ,c" ,d" ,e" and f". Join a'b'\b"c", 
c"d"4"e"/f"f"a". Also join o"a",o"b",o"c",o"d",o"e",o'f". 
o"-a"b"c"d"e'f" is the plan of the pyramid. In the hexagon 
a"b"c"d"e'f" the opposite sides should be parallel. The lines 
o"b",o"c",o"d",o"e",o'f",f"e",e"d"4"c",c"b" should be drawn 
full, and o"a",a"b'\a''f" should be drawn broken. This will 
be plain if we consider that the direction in which the object is 
seen is that indicated by the arrow. 

PLATE V. 

Prob. I. — To draw the plan and the elevation of a circle 
whose plane is parallel to the vertical plane, and at a distance of 
four inches from it. Diameter of circle^4". The elevation ot 
a circle, whose plane is parallel to the vertical plane, is a circle 
whose diameter is equal to that of the given circle ; and the 
plan is a straight line, whose length is equal to the diameter 
of the circle parallel to the ground line, at a distance from it 
equal to the distance from the plane of the circle to the vertical 
plane. This is illustrated in Prob. III. Plate II. Therefore 
assume a point ^ at a distance of three inches from the left 
side of the paper, and four inches from the ground line. 
With ^ as a centre, and a radius equal to two inches the radius 
of the given circle describe the circle abed. Draw a'd 
parallel to the ground line, four inches from it, the distance 
from the plane of the circle to the vertical plane. From a and 
c respectively the extremities of the horizontal diameter, draw 
a a' and c c' perpendicular to the ground line, intersecting a'c 
in the points a' and c' ; abed is the elevation and a' c' the 
plan of the circle. 

Prob. II. — To draw the plan and the elevation of a circle 
whose pla7ie is perpetidicnlar to the horizontal pla?ie, a7id which 
forms with the vertical pla7ie an angle of 45°. Diameter of 
circle=^^". The plan of a circle whose plane is perpendicular 



25 

to the horizontal plane will evidently be a straight line. This 
is illustrated in Prob. Ill, Plate II ; a'b' is the plan of one base 
of the cylinder, and a!' b" is the plan of the other base. The 
plane of each of these circles is perpendicular to the horizontal 
plane, and being parallel to the vertical plane its plan is 
drawn parallel to the ground line. The plan of a circle 
whose plane is perpendicular to the horizontal plane will 
be a straight line, forming with the ground an angle 
equal to that formed by the plane of the circle with the 
vertical plane. Therefore assume a point b' at a distance 
of ten inches from the left side of the paper, and of four 
inches from the ground line. Through b' draw a'c' form- 
ing an angle of 45° with the ground line. Lay off ^'<2'and 
b'c\ each equal to two inches, the radius of the circle, that 
is, make a'c' equal to four inches, the diameter, which is 
the plan of the circle. The elevation will be found as follows ; 
the point b' is the plan of that diameter which is perpendicular 
to the horizontal plane ; its elevation will therefore be a straight 
line perpendicular to the ground line, equal to the diameter. 
From b' draw b'b perpendicular to the ground line. Lay off 
from the ground line to 0, the elevation of the centre, a distance 
equal to four inches. From lay off ^ ^ and d, each equal to 
two inches, the radius of the circle, that is, make bd equal 
to the diameter. The line a'd which is the plan of the circle 
is also the plan of that diameter which is parallel to the hori- 
zontal plane. The elevation of this diameter will be parallel 
to the ground line. Therefore through o draw ac parallel to 
the ground line. From a' and d respectively draw a' a and dc 
perpendicular to the ground line, intersecting ac in the points a 
and c, the extremities of the elevation of the horizontal diam- 
eter. The elevation of the circle will be an ellipse, of which bd 
is the major axis and ac the minor axis. With a or ^ as a 
centre, and a radius equal to the semi-major axis or the 
radius of the circle, that is, two inches, describe an arc inter- 
secting the major axis in the points e and /, the foci of the 
ellipse. Having determined these points we may construct 
the curve as described in Chap. I, by the aid of the ellipso- 



26 

graph. The points e and / in Prob. II, Plate V, correspond 
with a and b in Fig. Ill, (Instruments), ab c d\'i>\h^ elevation 
required. 

Prob. III. — To draw the plan and the elevation of a right 
circular cylinder, whose axis is parallel to the horizontal platie^ 
and which forms with the vertical plane an angle of 45°. 
Diameter of cylinder^^.". Length^z^^''. Since the axis of 
the cylinder is parallel to the horizontal plane it will be pro- 
jected upon it in its true length, and its plan will form with 
the ground line an angle equal to that formed by the axis with 
the vertical plane, viz. 45". Also since the bases are perpen- 
dicular to the axis they are perpendicular to the horizontal 
plane, and will be projected upon it in straight lines, equal to 
the diameter of the cylinder, and perpendicular to the plan of 
the axis. Therefore assume a point b' at a distance of six 
inches from the right side of the paper, and of six inches 
from the ground line. From b' draw b' f five and one half 
inches long, the given length of the cylinder, forming an 
angle of 45° with the ground line. Through b' and/' re- 
spectively draw a' c' and c' g' , perpendicular to b' f, that is, 
forming with the ground line angles of 45^. Lay off b'a', 
b'C ,f'g' and/V,' each equal to two inches, the radius of the 
cylinder. Join a' e' and c' g' . a' c' g' e' is the plan of the 
cylinder, which is a rectangle, whose length is equal to that 
of the cylinder, and whose width is equal to the diameter 
The elevation will be determined as follows : Draw 00, parallel 
to the ground line, four inches from it. From b' and/' respec- 
tively draw b'o and /'^/perpendicular to the ground line, inter- 
secting 00, in and 0,. 00, is the elevation of the axis, and the 
points and 0, are the elevations of the centres of the bases. 
The ellipse, which is the elevation of the circumference of each 
base, will be constructed as in the preceding problem, bd equal 
to four inches, the diameter of the cylinder, will be the major 
axis, and ac the minor axis, determined by erecting perpendicu- 
lars to the ground line from the points a' and c', intersecting 
0, produced in a and c. Similarly the major axis / h and 
the minor axis e g are determined. The ellipses abed and 



27 

ef g h are the elevations of the bases. Draw bf and d Ji 
parallel to the ground line, a b f g h d is the elevation of the 
cylinder. The semi-ellipse/^ /^ is drawn broken, because it 
is invisible from the point of sight, the object being seen in 
the direction of the arrow. 

PLATE VI. 

Prob. I. — To draw the plan and the elevation of a circle, 
whose plane is parallel to the horizontal plane, and at a distaitce 
offonr inches from it. Diameter of circle=^" . It is illus- 
trated in Prob. Ill, Plate II, that the elevation of a circle 
whose plane' is parallel to the vertical plane is a circle, whose 
diameter is equal to that of the given circle ; also that the 
plan is a straight line, equal to the diameter, parallel to the 
ground line, at a distance from it equal to the distance from 
the plane of the circle to the vertical plane. It is therefore 
evident that the plan of a circle, whose plane is parallel to the 
horizontal plane, is a circle whose diameter is equal to that of 
the given circle ; also that the elevation is a straight line, 
equal to the diameter, parallel to the ground line, at a distance 
from it equal to the distance from the plane of the circle to 
the horizontal plane. Therefore assume a point ^ at a dis- 
tance of three inches from the left side of the paper, and 
of four inches from the ground line. With ^ as a centre, and 
with a radius equal to two inches, the radius of the given 
circle, describe the circle a'b'c'd'. Draw ac parallel to the 
ground line, four inches from it, equal to the distance from the 
plane of the circle to the horizontal plane. From a' and 
c' respectively, the extremities of the diameter which is par- 
allel to the vertical plane, draw a' a and c'c perpendicular to the 
ground line, intersecting ac in the points a and c. a'b'c'd' is 
the plan, and ac the elevation of the circle. 

Prob. II. — To draw the elevation and the plan of a circle, 
zv hose plane is perpendicular to the vertical plane, and which 
forms with the horizontal plane an angle of ^^°. Diameter of 
circle^^" . The plan of a circle, whose plane is perpendicular 
to the horizontal plane, is a straight line equal to the diameter, 



28 

forming with the groimd line an angle equal to that formed by 
the plane of the circle with the vertical plane. This is illus- 
trated in Prob. II, Plate V. Similarly the elevation of a circle, 
whose plane is perpendicular to the vertical plane is a straight 
line, equal to the diameter, forming with the ground line an 
angle equal to that formed by the plane of the circle with the 
horizontal plane. Therefore assume a point ^ at a distance 
of nine inches from the left side of the paper, and of four 
inches from the ground line. Through d draw ac, forming an 
angle of 45° with the ground line. Lay off da and dc, each 
equal to two inches, the radius of the circle, that is, make ac 
equal to four inches, the diameter, which is the elevation of 
the circle. The plan will be found as follows : The point d is 
the elevation of that diameter which is perpendicular to the 
vertical plane ; its plan will therefore be a straight line, per- 
pendicular to the ground line, equal to the diameter. From d 
draw dd' perpendicular to the ground line. Lay off from the 
ground line to 0, the plan of the centre, a distance equal to 
four inches. From lay off ob' and pd\ each equal to two 
inches, the radius of the circle, that is, make b' d' equal to the 
diameter. The line ac, which is the elevation of the circum- 
ference of the circle, is also the elevation of that diameter 
which is parallel to the vertical plane. The plan of this diam- 
eter will be parallel to the ground line. Therefore through 
draw a'c' parallel to the ground line. From a and c respect- 
ively draw aa' and cc' , perpendicular to the ground line, inter- 
secting a'c' in the points a' and c' , the extremities of the plan 
of the line. The plan of the circle will be an ellipse, of which 
b'd' is the major axis and a'c' the minor axis. The foci of the 
ellipse may now be determined as in Prob. II, Plate V, and 
the curve constructed as described in Chapter I, by the aid of 
the ellipsograph. 

Prob. III. — To draw the elevation and the plan of a right 
circular cone, whose axis is parallel to the vertical plane, and 
which forms with the horizojital plane an angle of 45°. Alti- 
tude of cone^6" . Diameter of base=4" . Since the axis of 
the cone is parallel to the vertical plane it will be projected 



29 

upon it in its true length, and its elevation will form with the 
ground line an angle equal to that formed by the axis with the 
horizontal plane, viz. 45°. Also since the base is perpendicu- 
lar to the axis it is perpendicular to the vertical plane, and will 
be projected upon it in a straight line, perpendicular to the 
elevation of the axis. Therefore assume a point a on the 
ground line, at a distance of four inches from the right 
side of the paper. From a draw ac forming an angle of 45° 
with the ground line, and make ac equal to four inches, the 
diameter of the base. Lay off a distance ab equal to two 
inches. From b draw bd forming an angle of 45° with the 
ground line ; that is draw bd perpendicular to ac. Lay off ba 
equal to six inches, the height of the cone. Join da and dc. 
dac is the elevation required. The plan will be found as 
follows : since the axis is parallel to the vertical plane it will be 
projected upon the horizontal plane in a straight line, parallel 
to the ground line. Therefore draw d'o parallel to the ground 
line, at a distance from it equal to four inches. From d and b 
respectively draw dd' and bb' , perpendicular to the ground line, 
intersecting d'o in d' and o. d'o is the plan of the axis, and 
the point is the plan of the centre of the base. The ellipse 
a'b'c'e', which is the plan of the circumference of the base, will 
be constructed as in the preceding problem, b'e' equal to four 
inches, the diameter of the base, will be the major axis, and 
a'c' the minor axis, determined by dropping perpendiculars 
from the points a and c, intersecting dc' in the points a' and c'. 
From d' draw d'g' and d'f'y each tangent to the ellipse. 
The lines d'g' and d'f are the extreme elements of the cone 
in the plan — the object being seen in the direction of the ver- 
tical arrow. The part of the ellipse f'a'g' will evidently be 
drawn broken, since it is invisible from the point of sight. In 
inking the ellipses in this Plate and in Plate V, a large portion 
of each curve may be drawn correctly enough for all practical 
purposes by arcs of circles. Theoretically this method of draw- 
ing an ellipse is incorrect ; but since arcs may be constructed 
which very nearly coincide with the curve for a considerable 
distance, it is desirable, in order to make all the parts symmet- 



30 

rical to draw them in this way. The construction is made as 
follows : — Let ab and cd (Fig. I.) be the major and minor 
axes of an ellipse. We will suppose the curve drawn" in 
pencil as described in Prob. II., Plate V. Draw two lines ek 
and eg (Fig. II.) forming with each other any angle, at the 
point e. With ^ as a centre, and a radius equal to the 
semi-major axis, viz.: oa or ob, describe the arc^ intersecting 
ek and eg respectively in the points /and^. Also with the 
same centre and a radius equal to the semi-minor axis, viz: oc 
or od, describe the arc hi intersecting ek and eg respectively in 
the points h*2cXi^ i. Join if, and through // and g respectively 
draw Jij and gk parallel to fi, intersecting eg in/ and ek in k. 
With a and b the vertices of the major axis as centres, and 
a radius equal to ej describe arcs, intersecting ab in 7i and 
/. With n and p as centres and the same radius describe 
arcs, apparently coinciding with the ellipse for a short distance. 
Also with c and d as centres, and a radius equal to ek 
describe arcs, intersecting cd produced in/ 3.ndm. With/ 
and 7n as centres and the same radius describe arcs, ap- 
parently coinciding with the ellipse for a short distance. Hav- 
ing determined how far the arc will coincide with the curve 
on one side of the axis it should be drawn an equal length on 
the other side. The four arcs may be connected by the aid of 
the curved ruler, and the student should make the joints care- 
fully, in order that the curve may appear continuous. 

The above construction gives the radius of curvature at the 
vertices of the major and minor axes. In order that a larger 
part of the curve may be represented by arcs, make the 
radius a little longer than na in the one case, and a little 
shorter than Id in the other. 



CHAPTER III. 

PRACTICAL PROBLEMS. 

Hitherto attention has been confined to geometrical prob- 
lems. The object has been, by means of simple figures, to 
illustrate some of the principles which are involved in making 
practical working drawings. A few constructions of this kind 
will now be given. 

PLATE VII. 

Problem. — To draw the plan and the elevation of a Pillow 
Block. — Draw aa' parallel and bb' perpendicular to the ground 
line. Each of these lines should be drawn indefinitely. From 
0, the point of intersection, lay off" oc and oc' , each equal to one 
inch and a half Through c and c' respectively, draw dd' and ee' 
parallel to the ground line. From ^ lay off ^(^, and ^^a^', each 
equal to six inches ; cf and cf, each equal to five-eighths of 
an inch ; 'ch and ch' , each equal to one inch and a half ; cj 
and cf , each equal to two inches and one-eighth. From the 
points ff',h,h'ff,d .and d' respectively draw fgf'g',hi,h'i\ 
Jk,jk', de and d'e' perpendicular to the ground line. With 
each of the points / and /' as a centre, and a radius equal 
to one inch, describe the semi-circles iwip and m'n' p' . From 
lay off oq and oq' , each equal to four inches and three- 
quarters ; also or and or\ each equal to four inches and one- 
eighth. With each of the points q,q\r and r' as a centre, and 
a radius equal to seven-sixteenths of an inch, describe a 
semicircle, limited in each case by a diameter drawn perpen- 
dicular to the ground line. Connect the extremities of the 
diameters drawn through q and r by lines parallel to the 
ground line ; also connect the extremities of the diameters 
drawn through the points q' and r' by lines parallel to the 
ground line, dd'e'e is the plan of the block. The elevation 



32 

will be found as follows : from e,e' ,Jt,n' ,k,k' ,i, and i' respect- 
ively draw ee^,e'e^^jm^,n'ii^^,kk^,k'k^^,u^ and i'i^^ perpendicular to 
the ground line. Lay off from the ground line to e, a distance 
equal to one inch ; to c^^ a distance equal to one inch and a 
half; to c^ a distance equal to four inches. From e^ draw e^e^/^ 
through c^^ dra.w g,or^/, and through c^ draw h^7i^^ parallel to the 
ground line. From ^ and ^ respectively draw^^^ ^^^£^'£',i 
perpendicular to the ground line, intersectrng^,^,/ in the points 
o-, and £-,,. Lay off i,s and z^^s^, each equal to one inch and five- 
eighths. Join so, and s,g;,. e,n,n,,e,^ is the elevation of the 
block. In inking the plan the following lines should be 
omitted, viz: the part of the linej'k between the points m and 
/. Also the part of the VmQ/k' between the points m' and/''. 
The centre lines aa' and dd' should be drawn in fine red ink ; 
also a short red ink line perpendicular to the ground line 
should be drawn through each of the points ^,r,//,r' and g', in 
order to locate the centres of the semicircles. In inking the ele- 
vation the line z,Zf, should be omitted. The perpendiculars to 
the ground line from the points n„k,,k,^ and n,, should be lim- 
ited by the line e,e,,; and the lines z,s and z,,s^ should be limited 
respectively at the points s and s,. The part of the line 
e,e,„ between the perpendiculars from k^ and k'\ should be 
omitted. The outlines of the holes rg and r' q\ being invisible, 
in the elevation, should be represented by broken lines. In 
order to complete the drawing all the dimensions should be 
added, a few of which are indicated in the plate. The student 
should be careful to write the figures neatly, a short distance 
from the line the length of which he wishes to indicate. As 
an illustration : in order to show the length of the line dd! , 
draw a fine red ink line parallel to it, limited by the lines 
ee^ and e'e^^, and write \2" as illustrated. Similarly the dis- 
tance between the perpendiculars 1,8 and i,^Si, viz., 3" is shown. 
To complete the drawing, the student should add all the 
dimensions which have been referred to, omitting the letters. 

EXERCISE III. 

To draw the plan of a square prism standing zipon a square 
pedestal, t/ie base of the pedestal forming with tlie horizontal 



33 ' 

plane an angle of 30°. Length of side of base of pedestal:='X," ; 
thickness of pedestal=k" • Length of side of base of prism= 
li" ; altitude of prisin^\\" . Assume a point a on the 
ground line. From a draw ab equal to throe inches, the < 
length of the side of the base of the pedestal, forming with 
the ground line an angle of 30°. From a a'"!d h respectively 
draw ac and <5,^ perpendicular to ab, that is, forming angles of 
60° with the ground line, each equal to. one-half inch, the 
thickness of the pedestal. Draw cd parallel to ab. Lay off 
ce equal to three-quarters of an inch, and of equal to two 
inches and a quarter. From e and/ respectively draw eg and 
fh perpendicular to cd, that is, forming angles of 60° with the 
ground line. Make eg equal to four inches and a half, and 
draw gh parallel to ab. abdc is the elevation of the pedestal, 
and ghfe is the elevation of the prism. Each of the lines 
gh, ef cd and ab represents two lines, parallel to the vertical 
plane ; ' the plans of these hnes will therefore be parallel to 
the ground line. Each of the points g, h,f?ix\d e is the eleva- 
tion of a line one inch and a half long, perpendicular to the 
vertical plane ; the plan of each of these lines will therefore 
be perpendicular to the ground line, one inch and a half long. 
And each of the points a, b, d and c, is the elevation of a line 
three inches long, perpendicular to the vertical plane ; the 
plan of each of these lines will therefore be perpendicular to 
the ground line, three inches long. It should be understood 
that the centre of the lower base of the prism coincides with 
the centre of the upper base of the pedestal. With these data 
the student is required to construct the plan. 



34 



PLATE VIII. 

Problems. To draw the elevation and the plan of a pedestal 
in two positions. Draw ak' (No. i) perpendicular to the 
ground line. Draw bbii,cc,„dtd^, and e^e,, parallel to the 
ground line, at distances from it respectively equal to five 
inches and a quarter, four inches and fifteen sixteenths, four 
inches and five eighths, and five eighths of an inch. Lay off 
ab, and ab,, each equal to two inches and five-sixteenths, 
and from b, and b^^ respectively, draw bj, and b^J^^ perpen- 
dicular to the ground line, intersecting d^d^^ in d, and d^^ ; e^e,^ 
in e, and e^^ ; and the ground line in/, and/^^. From o lay off 
oCf and oc,, each equal to one inch and one-eighth, and draw 
Cig, and c,,g,t perpendicular to the ground line. With o as a 
centre, and radii respectively equal to thirteen-sixteenths 
of an inch, and one inch and seven-sixteenths, describe the 
semicircles h,hti and j,kj^i. Draw l,m, and l,,m^f each at a 
distance of one-quarter of an inch from the centre line ak, per- 
pendicular to the ground line. Connect the semicircle j,kj,, 
with the straight lines d,djiJ,inj and l,^m,, by tangent arcs, 
described with a radius equal to three-eighths of an inch. 
Also connect the perpendiculars l/in^ and lf,m,, with the hori- 
zontal line e^e^i by tangent arcs, described with the same radius^ 
In order to find the centres of these arcs, with the centre o 
and a radius equal to one inch and thirteen-sixteenths. 
(that is, three-eighths of an inch longer than the radius with 
which we describe the semicircle j\kj\,), describe an arc o^o^,. 
Draw a line parallel to d,d,, at a distance of three-eighths of 
an inch from it, intersecting the arc O/O,, in a point o/, and a 
line parallel to l^,m,, at a distance ot three-eighths of an inch 
from it, intersecting the arc 0,0,, in a point o,^. Also draw a 
line parallel to e,et, at a distance of three-eighths of an inch 



35 

from it, intersecting OnOm in the point o^,,. With each of the 
points 0^,0, t and o,^^ as a centre, and a radius equal to 
three-eighths of an inch, describe an arc, which will evidently 
connect the lines as required. Similar constructions should 
be made on the other side of the figure. From the points 
d, and d,, respectively, lay off d,n, and di,n,^y each equal to 
five-eighths of an inch. Connect 7i,, and e^, by the arc 
of a circle, described with a radius equal to four inches. 
The centre of this arc will be found as follows : with n,f as 
a centre and a radius equal to four inches, describe an arc ; 
also with 61, as a centre, and the same radius, describe 
another arc, intersecting the former in a point 0' . With 
this point as a centre, and a radius equal to four inches, describe 
the arc n^,eii. A similar construction should be made 
in order to connect the points n, and e,. bibufnf, is the 
elevation of the pedestal. 

The plan will be drawn as follows : draw b/b and b,ib' perpen- 
dicular to the ground line. Lay of^ pq and pb, respectively 
equal to five-eighths of an inch, and three inches. Dv2iW pp' ,qq' 
and bb' parallel to the ground line. Prolong the lines l^m^ and 
l,,m,, respectively, to / and /, and connect these lines with 
qq' by tangent arcs, described with a radius equal to three- 
eighths of an inch. Find the centres of these arcs by the method 
described for the point 0,,, in the elevation. From the points 
g„h,,^ii and^// respectively, dvdiw g,g,k,k,h,,h' Andgng' perpen- 
dicular to the ground line, pp'b'b is the plan of the pedestal. 
To draw the plan and elevation in another position : assume 
the point b (No. II.) and from it draw bb' equal to bb' (No. I.) 
forming an angle of 30° with the ground line. From b and 
b' respectively, draw bp and b'p' perpendicular to bb' , equal to 
the corresponding lines in No. I, and connect/ with/'. Make 
pq, No. II., equal to pq, No. I., and draw qq' parallel to //'. 
Also lay off bg,bh,bl, etc., respectively equal to bg,bh,bl, etc.. 
No. I., and through the points gyh,l, etc., draw parallels to bp. 
The plan No. II. is a copy of No. I., the plane, //^ forming 
with the vertical plane an angle of 30°. 



36 

The elevation, No. II, will be found as follows : prolong the 
lines bfi^^, c^c^^, d^d,, and £,6,, (No. I.) indefinitely. Erect per- 
pendiculars from each of the points /, b, g, /, /', g', g" and b\ 
No. II, limiting them as indicated in the plate, that is, the 
perpendiculars from p, b and //, will be limited between the 
ground line and e^e,,, and between did,, and b,b,, ; the per- 
pendiculars from / and /' will be limited between the line 
6,61, and the semi-ellipse, which is the projection of the semi- 
circle jikj ,, ; the perpendiculars from g, g' and g" will be 
limited between the lines b,b,, and c^c,,.. The curve 71, e„ No. 
I, represents two curves, of which rnp and r'jt'q are the plans. 
Assume any point r, on the curve fi^e^. From 71^ and r, draw 
perpendiculars to the ground line, intersecting pp' and qq' in 
r, 7^, 71 and 71' . Lay off p7i and pr, No. II, respectively equal 
to p7i said pr, No. I. Draw ;^;/ and r/ parallel to pq, No. II. 
From 71 and ;/, No. II, respectively, draw 7i7i,,, and 7i'7i, per- 
pendicular to the ground line, intersecting d,,d, in 7t^^^ and 
71^. From r , No. I, draw a parallel to the ground line, and 
from r and r', No. II; erect perpendiculars intersecting this 
line in r^^ and r^. From/ and q, No. II, erect perpendiculars 
intersecting e,e,, in e and /. Through the points 71^^^, r^^ and 
e pass a curve. Also through the points 7^^, r^ and / pass a 
curve. These curves should be drawn with the curved ruler, 
which should be fitted to the points, any number of which 
may be found in a manner similar to that described for 
r^^ and r^. The ellipses /i^zA^j and 7\^j]^ will be found as fol- 
lows : from k', No. II, erect a perpendicular, intersecting 
j\j\^ in 0, the centre of the curves. From /i and /i' erect per- 
pendiculars, intersecting j)j\^ in /i^ and k^^. From the point 
i, No. I, draw a parallel to the ground line, intersecting k'o, 
No. II, in z. oi is the semi-major axis, and hjt^^ is the minor 
axis of the smaller ellipse. On these axes the semi-ellipse 
hfih^^ may be constructed, as in Prob. II, Plate V. From 
j^ and /^^, No. I, draw perpendiculars intersecting bb' in j and 
f . Lay off k'j and k'f, No. II, each equal to k'j or k'j\ No_ 
I. From j and j' , No. II, erecL perpendiculars, intersecting 
j,j,i 'vcij, a.ndj\^, the vertices of the minor axis of the larger 



ellipse. The vertex of the major axis k will be found by 
drawing from k, No. I, a line parallel to the ground line, in- 
tersecting k'o, No. II, in k. It will be observed that the hori- 
zontal lines d^d^^ and the vertical lines Ifin^ and l,;m^^ are 
connected with the larger ellipse by short curves. Also that 
the horizontal line 6^6^^ is connected with the vertical lines 
l^ryif and /^^m^^ by short curves. The construction of these 
curves, as well as the visible portion of the curve sty is left 
as an exercise for the student, ebfi^ff^^ is the elevation re- 
quired. 

PLATE IX. 

Problems. To draw the plan and the elevation of a crank in 
two positions. Draw aa' , No. I, perpendicular to the ground 
line. Lay off from the ground line to o a. distance equal to 
two inches and eleven-sixteenths. With ^ as a centre, and 
radii respectively equal to two inches and eleven-six- 
teenths, two inches and a half, and one inch and nine- 
sixteenths, describe the circles d,dj^, C/C,, and d^d^,. Lay off 
from the centre of the circles along aa' a distance equal to one 
inch and seven-eighths, and draw e^e,, parallel to the ground 
line. Lay off from aa' to e^ and e,^ distances each equal to 
seven-sixteenths of an inch, that is, make e^e^^ equal to seven- 
eighths of an inch. From e, and e,, draw perpendiculars to 
the ground line, and produce them until they intersect the 
circle d^d^, in the points // and f^. Lay off from ^ to ^^ a 
distance equal to ten inches. With o, as a centre, and 
radii respectively equal to one inch and five-eighths, one inch 
and a half, fifteen-sixteenths of an inch and eleven-sixteenths 
of an inch, describe the circles gtgiph,h^f,i^i^^ andJJ^i. Draw 
k,k^, parallel to the ground line and tangent to the circle g^g^,, 
and lay off from aa' to k^ and k^^ distances each equal to one 
inch and a quarter. Also draw 1,1^^ parallel to the ground line 
and tangent to the circle b^b^^, and lay off from aa' to /, and 
/^^ distances each equal to one inch and three-quarters. Draw 
the straight lines k^l^ and k,,l^i, and connect each of them with 
the circle gg^, by a tangent arc, described with a radius equal 



38 

to one inch. Also connect each of the lines k,l, and k,,l^^ with 
the circle b,b,, by a tangent arc, described with a radius equal 
to two inches and a half. The constructions for finding the 
centres of these tangent arcs are illustrated, and are similar 
to those employed in No. I., Plate VIII., to which the student's 
attention is directed, gignbub, is the elevation of the crank. 

The plan will be found as follows : Draw cc' parallel to the 
ground line, and on aa! mark points at the following dis- 
tances from cc\ viz.,TV',F, t", if",2f", 2^,4^' and 5". 
Through these points, respectively, draw mm' ,hh' ,mt' yUu' ,bb' , 
pp',qq' and ii' parallel to the ground line. The length of these 
lines will be determined as follows : tangent to each circle in 
the elevation, draw two lines perpendicular to the ground line, 
intersecting the plan in its extremities. A careful examination 
of the Plate will enable the student to see clearly the connec- 
tion between the plan and elevation of each circle. The 
line j is connected with the points p and q by quadrants, 
described with a radius equal to one-quarter of an inch, and 
the centres of these quadrants are on the line pq. Similarly 
the liney, is connected with the points /' and q'. The points 
c and m are connected by a quadrant, described with a radius 
equal to three-sixteenths of an inch, and the centre of this 
quadrant is at the intersection of the lines bin and c'c. 
The quadrant c'm' is constructed in a similar manner. The 
points h and 71 are connected by a quadrant, described with 
a radius equal to one-eighth of an inch, and the centre of 
this quadrant is at the intersection of the lines gii and h'h. 
The quadrant h'n' is drawn similarly. With 0' , the point of 
intersection of pp' with aa', as a centre, and a radius 
o'i describe the arc ii' . From e, and e,, draw perpendiculars 
intersecting cc' m t and f , and bb' in e and e' . cc'i'i is the plan 
of the crank. 

Draw the plan and elevation in another position as follows : 
make the plan, No. II, a copy of the plan No. I, 
and make those lines, which in No. I. are parallel and perpen- 
dicular to the ground line, form angles of 45° with the ground 
line in No. II. 



39 

The elevation will be found as follows : from o^, No. I, draw 
0^0^ parallel to the ground line. This line will contain the 
centres of the ellipses which are the projections of the circles 
ii', qq\ pp' , ss' , gg' and 7tn'. Also from o, draw oo parallel to 
the ground line. This line will contain the centres of the 
ellipses which are the projections of the circles bb\ mm' ^ 
dd' and rr' . The major axis of the ellipse will, in each case, 
be perpendicular to the ground line, and equal to the diameter 
of the circle of which the ellipse is the projection. It is 
needless to give the construction for each one, as the method 
of making it is illustrated in Prob. II, Plate V. The points 
kf.kn^lf and /^^, No. I, will in the plan be found in the line bb' . 
Transferring these points by measurement to the corresponding 
position in No. II, their elevations will be found as illustrated. 
Join kf If and k,, /^. From k draw kv perpendicular to bb\ in- 
tersecting uu in V. From v draw vv^ perpendicular to the 
ground line,- intersecting kk^^ in v ^. Through v^ draw vp^ 
parallel to kl^. From e^ and /^, No. I, draw ee^^ and fj^^ par- 
allel to the ground line. From the points e and / in the plan 
No. II, draw perpendiculars to the ground line, intersecting e^^^ 
in e^ and e^^, d.n^fj^^ in /^ and/^. e^ ^„f,if, is the visible part 
of the key-way. The student should construct the tangent 
curves connecting the straight lines kl^ and k^l^^ with the 
ellipses. 

Ex. IV. — To draw the plan of an hexagonal pjdsm standing 
2pon an octagonal prism ; the planes of the bases forming 
with the horizontal plane angles of 30°, Let the di- 
mensions and position of the solid, relative to the vertical 
plane, be the same as in Prob. IV, Plate I. That is, assume 
the elevation eiq'v" a copy of eiq"v", Plate I. In Plate I. the 
planes of the bases are horizontal ; hence, their elevations are 
parallel to the ground line. In the exercise they will form 
with the ground line angles of 30°. Also in Plate I. the 
elevations of the edges, which are perpendicular to the bases, 
are perpendicular to the ground line. In the exercise they 
will form with the ground line angles of 60°. If it be 
remembered that in the exercise the position of the figure 



40 

relative to the vertical plane is the same as that in Plate I., 
it will be apparent the plans of the edges, which are perpen- 
dicular to the bases, will be parallel to the ground line, and 
their distances from it will be found in the plan, Plate I. 
With these data the student is required to make the con- 
struction. 

PLATE X. 

Problems. To draw the elevation and the pla7i of a crank 
shaft in two positions. First draw the elevation No. I. when 
the axes of the shaft and crank-pin are parallel to the vertical 
plane, the crank being in a vertical position. Let the diameter 
of the shaft be two inches, the diameter of the pin one inch 
and three-quarters, and the distance between the axes of the 
shaft and the pin five inches. Draw aa, the elevation of the 
axis parallel to the ground line, at a distance from it equal to 
one inch, that is, one-half the diameter of the shaft. Draw 
bb, the elevation of the axis of the pin parallel to the ground 
line, at a distance from aa, equal to five inches. Draw 
cci perpendicular to the ground line. Draw dg parallel 
to the ground line, at a distance of one inch from bb,. From 
c, the point in which dg intersects cc,, lay o^ cd equal to one 
inch and three-quarters, of equal to two inches and a half, and 
eg equal to four inches and a quarter. From d and g draw 
perpendiculars to the ground line, limited by the line aa, re- 
spectively in the points d , and^,. Also from/ draw^, per- 
pendicular to the ground line. 

Draw hj parallel to the ground line, at a distance from it 
equal to two inches, the diameter of the shaft. On hj from 
dd, lay off" to h a distance equal to three inches, and draw JlJl^ 
perpendicular to the ground line, also on /^/'from gg, lay off 
to/ a distance equal to three inches, and draw jj^ perpendicu- 
lar to the ground line. From o, the point in which bb, inter- 
sects cc, lay off oe and oe, each equal to seven-eighths of an 
inch, that is, one half the diameter of the pin, and draw ei and 
e,i, parallel to the ground line, h^dgj^ is the required elevation, 



41 

To construct the plan, draw a' a" parallel to the ground line. 
Draw kl and k'l\ parallel to a'a" , each at a distance from it 
equal to one inch, that is, one half the diameter of the shaft. 
Produce the perpendiculars hh^,dd„cc^,ffiqq^ and jj\ intersect- 
ing the lines kl and kj, in the points k^k' ,m,in\nyn' ,p,p' ,q, q',l, 
and /'. Draw rs and r's' parallel to a'a", each at a distance 
from it equal to seven-eighths of an inch, that is, one-half the 
diameter of the crank pin. kll'k' is the required plan. 

Now draw the plan and elevation (No. 11.) when the axes 
of the shaft and pin are parallel to the horizontal plane, and 
form with the vertical plane angles of 30°, the crank being in 
a vertical position. Draw the plan, making it a copy of the 
plan No. I, drawing all those lines which in No. I are parallel 
to the ground line, forming angles of 30° with the ground line ; 
also all those lines which in No. I are perpendicular to the 
ground line, forming angles of 60° with the ground line, that 
is, make one set of lines perpendicular to the other set. 

In drawing the elevation, we suppose that the shaft is rest- 
ing upon the horizontal plane as in No. I. The axes of the 
shaft and pin will therefore be at the same distance from the 
horizontal plane as in No. I, that is, their elevations will be 
in the prolongations of these lines, as illustrated in the plate. 
These broken lines will contain all the centres of the ellipses 
which are the projections of the circles. The major axes of 
the ellipses will be perpendicular to the ground line, drawn 
from the points in the plan in which the plan of the axes 
intersects the lines which are perpendicular to it, and in each 
case the major axis will be equal to the diameter of the circle 
ol which the ellipse is the projection. With these suggestions, 
and with the aid of the illustration, the student will have 
little difficulty in constructing the elevation. 



42 



CHAPTER IV. 

ISOMETRIC DRAWING. 
PLATE XI: 



A series of exercises on Isometric Drawing, sometimes 
known as Isometric Perspective, will now be given. By this kind 
of drawing a view of the object to be represented is produced, 
which gives to the observer a better idea of its form than that 
which is suggested by the plan and elevation. 

The simplest way to approach the subject will be by 
making a projection of a cube as follows : assume a point o, 
Fig. I, and draw oa of any convenient length, perpendicular 
to the ground line G.L. With ^ as a centre, and the 
radius oa, describe the circle abed e f, and within this circle 
inscribe the regular hexagon d b c def. Join oc and oe. The 
lines ed, oc and ab are parallel and equal to each other ; the 
lines cd, oe and af are parallel and equal to each other ; and 
the lines cb, oa and ef are parallel and equal to each other. 
From b, d and / draw broken hues respectively parallel to 
af, ef2Lnd ab, intersecting at the point o. Erase the circle, 
and we shall have the projection of a cube when a great diag- 
onal is projected in the point o, that is, when this diagonal is 
perpendicular to the vertical plane, all the edges being pro- 
jected in lines of equal length. Such a representation of a 
cube is called its isometric projection. Since the great diag- 
onal is perpendicular to the vertical plane, the adjacent 
edges oa, oc and oe form with the vertical plane equal angles, 
and are projected in lines of equal length, as we have illus- 
trated. Also, since the length of the edges oa, oc and ot are 
equal, the points a, c and e are equidistant from the vertical 
plane, and the lines joining these points will be parallel to the 
vertical plane, and will be projected upon it in their true 
length. Join ac, ce and ea. Now, since the line ec is pro- 
jected in its true length, it is the length of the diagonal of the 



43 

square which is projected in the parallelogram ocde. In order 
to find the true length of the side of this square, from the points 
c and e draw straight lines, forming with ec angles of 45°, inter- 
secting in the point d' . Similar lines constructed on the other 
side of the diagonal would complete the square, but for the 
purpose the lines which are drawn are all that are necessary. 

To construct the projections of the circle inscribed in the 
visible faces of the cube, the major axis of the ellipse in 
each case will be the projection of that diameter of the circle 
which is parallel to the vertical plane, and will be contained 
in one or other of the lines connecting the points a, c and e. 
Now, since the diameter of the inscribed circle will be equal 
to the length of the side of the square, the axes will be found 
as follows : bisect ac in h and lay off hi and M, each equal to 
one-half the length of the line cd' or ed' ; tk will be the major 
axis of the ellipse. In order to construct the minor axis, con- 
ceive that a square is inscribed within the circle, and that i 
and k are the projections of two of the angular points of this 
square. Draw il and km parallel to oc, and draw im and kl 
parallel to oa. il intersects in kl in /, and im intersects km in 
m. Join Im, which will be perpendicular to ik. Im. will be the 
minor axis of the ellipse. On ik and Im, as the axes, con- 
struct the ellipse i m k I, which will be tangent to the lines 
oc, cb, ba and ao at their middle points. Similarly construct 
the ellipses, which are the projections of the circles inscribed 
in the faces ocde and ef a. It will be observed that these 
ellipses are equal to each other in every respect, that the 
major axes produced form the equilateral triangle e c a, that 
the line ec is parallel to the ground line, and that the lines ea 
and ca form with the ground line angles of 60°. These three 
ellipses are the projections of circles in the different positions 
which they generally occupy in practical problems. The illus- 
trations have therefore been given complete for future reference. 

In the drawing there are three sets of parallel lines, one 
set being perpendicular to the ground line, the remaining 
lines forming with the ground line angles of 30°. A draw- 
ing of this kind may be made very conveniently with the 



44 

aid of the 30° triangle applied to the T square. Hereafter it 
will be unnecessary to draw the ground line which has been 
introduced for reference. It will be understood that perpen- 
dicular lines are drawn perpendicular to the edge of the blade 
of the T square, and that 30° lines form with this edge angles 
of .30°; that the major axis of an ellipse will be ruled along 
this edge, as in the ellipse inscribed in the face of the cube, 
ocdey or that the major axis will form with this edge an angle 
of 60°, as in either of the other cases. 

In the illustration that has been given we have not been con- 
fined to any particular dimensions. But if it be required to 
draw the isometric projection of a cube of any given dimensions, 
it will be necessary to construct a scale from which to take 
measurements. It is evident in Fig. I that each edge of the cube 
is projected in a line which is shorter than its true length. 
Now, if we can ascertain the true length of the edge which is 
projected in the line cd, we may then construct a scale whose 
length shall bear to the length of the ordinary scale the same 
proportion which the line cd bears to the line of which it is the 
projection. The length of this line will be found as follows : 
We have seen that the true length of the diagonal of the face 
is projected in the line ec, and that the lines cd' and ed' , form- 
ing angles of 45° with the diagonal, determine two sides of 
the square. Conceive this square to be revolved about its 
diagonal ec into a position when the point d' is projected in d- 
In that position the line cd' is projected in cd, and the line 
ed' in ed. Hence, cd' is the true length of the line of which 
cd is the projection, and since the line joining d' with d is 
perpendicular to ec, it is evident that if we lay off from c to 
s' a distance equal to two inches, and from s' draw s' s perpen- 
dicular to ec, intersecting cd in s, and cs, will be the projection 
of a line two inches long ; that is, when the square is re- 
volved into the position ocde, the point s' will be projected in 
s. From this it will be plain that the following will be a cor- 
rect construction of the required scale : at any point a, Fig. 
II, draw ab and ab' forming with ac respectively, angles of 
45° and 30°. Lay off ab equal to the length of the 



45 

required scale, we will say twelve inches. From b draw be 
perpendicular to ac, intersecting ab' in the point b' . ab' will 
be the length of the required scale. In order to compare this 
construction with that of Fig. I, complete the square abed, 
the length of whose side is twelve inches. If we conceive 
this square to be revolved about its diagonal ac until the sides 
are projected in the lines ab' fi'c^cd' and d'a, which form with 
ac angles of 30°, we shall have the projections of the sides of a 
square twelve inches long, and the length of either of these 
lines will be the length of the required scale. A scale which 
bears to the ordinary scale the same proportion which 
ab' bears X.o ab is called an isometric scale. We have divided 
the ordinary scale cb into twelve equal parts, that is, into 
twelve inches, and the corresponding points of division on the 
isometric scale cb' are determined by drawing perpendiculars 
to the line ac, as illustrated. 

PLATE XIT. 

It has been shown (Fig. II., Plate XI.) how to find the 
length of an isometric scale. 

Before drawing the problems of this plate, the student 
should construct an ordinary scale, No. I, twelve inches long, 
and also an isometric scale. No. 2, of the same length, that is, 
the length of No i should be made equal to that of be — Fig. 
II., Plate XL, and the length of No. II., should be made equal 
to that of b'c in the same figure. It is understood that these 
constructions are to be drawn full size. Having made these 
scales, we shall be ready to construct the following: 

Prob. I. — To draw the isometric projection of a rectangular 
block, six inches long, three inches wide, and one inch thick. 

The elevation of a block of these dimensions is a rectangle 
abed, whose length ab is equal to six inches, the given length, 
and ad equal to one inch, the thickness of the block. The 
plan is a rectangle a'b'e'f , whose width a'f is equal to three 
inches, the given width. 

To draw the isometric projection, assume a point a„ and 
draw a perpendicular a^a^d^. Also draw 30° lines a,b, and a,f, ; 



46 

a^d^ will be perpendicular to the edge of the T square, and 
a^b^ and aif, will form angles of 30° with this edge, that is, 
these lines will form with each other angles of 120°. Now it 
is evident that if we make aid, one inch long by the shorter 
or isometric scale, and lay ofif a,gi and a,h, each equal to a/d, 
and complete the regular hexagon d,h,g,, as in Fig. I., Plate 
XL, we shall have the isometric projection of a cube the 
length of whose edge is one inch ; and it is also evident that if 
we prolong the line a,g, to b, making the length of afi, equal 
to six times the length of a^g^, that is, equal to six inches by 
the isometric scale, we shall have the isometric projection of 
a line six inches long ; and also, if we prolong a^h, tof„ making 
the length of a,f, equal to three times the length of a^/i^, that 
is, equal to three inches by the isometric scale, we shall have 
the isometric projection of a line three inches long ; and since 
a,d, is the isometric projection of a line one inch in length, and 
the remaining edges of the block are parallel to one or other 
of the lines a,d^,a,b, or a,f^, the following construction will com- 
plete the drawing : from b, and/^, respectively, draw perpendic- 
ulars b,c^ and///, that is, parallel to a^d/, from/ and d, respect- 
ively, draw//^/ and d,c^ parallel to a^b,; and from ^^and d,, re- 
spectively, draw b^e^ and dj, parallel to a^f,. f,e,b,c,d,lf is the 
required projection. 

The construction may be described in a few words as follQws : 
Assume a point a^. From a^ draw the perpendicular 
a^dt and the 30° lines a,bi and a J, indefinitely ; lay off 
by the isometric scale a^d^ equal to one inch, a^f^ equal 
to three inches, and afi, equal to six inches. Com- 
plete the figure by drawing parallels to these lines, from 
the points d„b^ and /. The broken lines represent the 
invisible edges. 

Prob. II. — To draw the isometric projection of a block whose 
plan a7td elevation are given. — Assume the plan as follows : 
draw a'b' and a'c' perpendicular to each other, respectively 
equal to five inches and three-eighths, and to three inches. 
Lay off a'd' equal to three inches and a half, and draw 
d'e' perpendicular to a'b' , equal to two inches and five-eighths. 



47 

Join b'e' by a straight line, and connect the points c' and e' by 
the arc of a circle described with a radius equal to four inches. 
a'b'e'c' is the plan of the solid. The elevation is the rectangle 
abfg, the length ab being equal a'b', and ag being equal to one 
inch and a quarter. 

To draw the isometric projection, assume a point a^ and 
draw by the isometric scale the 30° lines afi^ and a^c^ re- 
spectively, equal to five inches and three-eighths, and three 
inches, and the perpendicular a^g^ equal to one inch and a 
quarter. Draw the y° lines gj^ and gji^ and the perpen- 
diculars b^f^ and c,h,. Lay off a^d^ equal to three inches and 
a half by the isometric scale, and draw the 30° line d^e^ equal 
to two inches and five-eighths. Join b^e^. The projection 
of the curve c'e' will be found as follows : assume a series 
of points, i',k', &c., &c., upon it. From these points 
draw perpendiculars to a'b', viz. : i'i" and k'k", Slc, &c. 
Measure the distance a'l" by the ordinary scale, and lay off" 
this distance taken from the isometric scale from a^ to i^^. 
From z,^ draw a 30° line. Measure the distance i"i' by the 
ordinary scale, and lay ofi" this distance taken from the 
isometric scale from i^^ to z/, i^ is one point of the curve. 
Similarly find the point k^ and having found a sufficient num- 
ber of points rule a curve c^i^k^e^ through them. The number 
of points which it is necessary to construct depends entirely 
upon the length of the curve whose projection is required. It 
is not difficult to decide the number, and this must be left to 
the discretion of the draughtsman in any given drawing. We 
have not drawn the invisible lines as in the preceding problem, 
in order not to confuse the illustrations which have been given 
for constructing the points ; but after drawing the upper 
curve the construction lines may be erased, and the lower 
curve and straight lines may be drawn. 

All the straight lines in the cube. Fig. T, Plate XI, are called 
isometric lines ; on any one of them a dimension may be laid 
off by the isometric scale. Also all the lines in the isometric 
projection, Prob. I., Plate XII, are isometric lines, every line 
being parallel to an edge of the cube. Now in Prob II. of this 



48 

plate, while the Unes a/b^,a^c^,a^g^, and all the lines which are 
parallel to one or other of these, are isometric lines, those 
which are not parallel to one or other of these, as b^e^, are 
called non-isometric lines. It must be distinctly understood 
that, in constructing an isometric projection, we must first 
ascertain the dimensions of the object to be represented by 
the ordinary scale, and these measurements must be made in 
three directions perpendicular to each other ; we then lay off 
these dimensions, taken from the isometric scale, on lines 
forming with each other angles of 120° as we have illustrated ; 
but in order to draw the projection of a non-isometric line, as 
b'e' , one or both of its extremities must be located by iso- 
metric lines. 

PLATE XIII. 

Prob. I. — To draw the isometric p7VJection of a pyramid 
standing upon a pedestal. — The dimensions may be assumed 
as follows : — draw the rectangle a'b'e'd', the plan of the 
pedestal, four inches long by two inches wide, and the rect- 
angle abef, the elevation of the pedestal, four inches long by 
one-half inch. The plan of the base of the pyramid will be 
the rectangle g'h'i'k' , three inches long by one inch and a half 
wide, the distance between the lines a'd' and k'g' being equal 
to one-half inch, and the distance between the lines a'b'2iVi^ 
o'h' being equal to one-quarter of an inch. To assume the plan 
of the vertex 0' , lay off g'l' equal to two inches, draw I'o' per- 
pendicular to^ 'h' equal to one inch. Join o'g,o'h',o'i' and o'k'. 
These lines are the plans of the edges of the pyramid. The ele- 
vation of the pyramid will be found by projecting the points^ and 
h' respectively in g and h, and the point 0' in 0, marking it at 
a distance of two inches and a half from ab, this being the 
altitude of the pyramid. To draw the isorhetric projection : 
assume a point a^ and construct the projection ol the pedestal 
as in Prob. I., Plate XII., the dimensions being four inches by 
two inches by one-half inch, and each of these dimensions will 
be laid off from the isometric scale. It must be remem- 
bered that all dimensions on isometric li7ies shoidd be 



49 

laid off from the isometric scale. In order to find one point, 
g^, of the base of the pyramid, produce k' g' in the plan to m' . 
cim' is equal to one-half inch, and g'm' is equal to one quarter 
of an inch. Make a,m, equal to one-half inch, and draw the 
30° Hne m^g, equal to one quarter of an inch. Draw the pro- 
jection of the rectangle three inches by one inch and a half, 
the sides of which will be 30° lines, as illustrated. To find 
the projection of the vertex lay off ^// equal to two inches, and 
draw the 30° line l^o^ equal to one inch. These distances 
correspond with those in the plan. From 0^ draw the per- 
pendicular 0^0^^ equal to two inches and a half, which cor- 
responds with the altitude given in the elevation. Join o^, 
with each of the angular points of the base of the pyramid. 
o^fi^f^d, is the required projection. 

Prob. II. — To draw the isometric projection of an upright 
crosz. The dimensions may be assumed as follows : draw the 
rectangle a'b'c'd' , the plan of the cross, three inches and a 
quarter long by one-half inch. The elevation of this part will 
be the rectangle abef, three inches aud a quarter long by three- 
quarters of an inch. Lay off ag equal to one inch and a 
quarter. Draw gh perpendicular to ab one inch and a quarter 
long, and complete the rectangle hikl, four inches and a half 
by three-quarters of an inch. The lines in the plan, m'h' and 
n'i\ are respectively the prolongation of the lines hi and ik- 
To draw the isometric projection : assume a point a^, and con- 
struct the projection of the cross piece as in Prob. I., Plate 
XII., the dimensions being three inches and a quarter by one- 
half inch by three-quarters of an inch. An examination of 
the drawing will show on which line each dimension is laid 
off. Lay off a^g^ equal to one inch and a quarter and draw 
the perpendicular gih^ equal to one inch and a quarter. 
Taking h^ as the projection of the angular point H of the 
rectangle h'i'n'rn in the plan, draw the projection of this 
rectangle three-quarters of an inch by one-half inch, and from 
each of the angular points draw a perpendicular four inches 
and a half long, the altitude of the cross, which is shown in 
the elevation. Connect the extremities of these perpendicu- 



50 

lars by lines which form the projection of the lower end of the 
upright piece, that is, of a rectangle three-quarters of an inch 
by one-half inch, nbk^d, is the required projection. The 
visible edges are drawn full, and the invisible edges broken. 
It will be observed that all the lines in this figure are isometric 
lines. 

In Prob. I. all the lines which are the projections of the 
edges of the pedestal, and of the sides of the base of the pyra- 
mid are isometric lines, while those which are the projections 
of the edges connecting the vertex with the angular points of 
the base are non-isometric lines. 

The attention of the student is particularly called to the 
construction for finding the projection of the vertex, as it is 
the key to the whole subject of isometric projection, o,, is 
found by means of three measurements, viz.: gl^^lp^ and 
o^Oii, the situation of this point being located by the plan 
o' and the elevation o^ from each of which we can take two 
measurements. 

Ex. V. — To draw the elevation of a hollow cylinder whose 
axis is parallel to the horizontal plane, and which forms with 
the vertical plane an angle of 45°. Let the rectangle a'b'c'd' , 
six inches long by three inches and a half, be the plan of the 
cylinder. The sides of this rectangle form with the ground 
line angles of 45°. Draw e' f parallel to a'b' , at a distance 
from it equal to one-half inch. Also draw g'h parallel to 
c'd' , at a distance from it equal to one-half inch, and draw 
i'k , the plan of the axis. The lines a'd' and b'c' are the 
plans of the bases of the cylinder, and the distance between 
the broken lines e' f and g'h' measures the diameter of the 
hole, the thickness of the material being one-half inch. 

PLATE XIV. 

Problem. — To draw the isometric projection of a solid bounded 
by plane and curved surfaces. Let the plan and elevation of the 
solid be assumed as follows : draw the rectangle a'b'c'd', the 
plan of the pedestal, four inches by two inches. Also draw 
the rectangle e'f'g'h', the plan of the upper horizontal plane 



51 

surface, one inch by one quarter of an inch. The sides e' f 
and g'h' should be parallel to a'b' and c'd! , and at distances 
from these lines respectively of seven-eighths of an inch. 
Also eh' and f g' should be parallel to a'd' and b'c' , and at 
distances from these lines respectively of one inch and a half. 
Join e'a',fb',g'c' and h' d' . e'f'g'h'—a'b'cd' is the plan of 
the solid. To construct the elevation : draw the rectangle abik 
four inches by one-half inch ; draw ef ono. inch long parallel to 
ki, at a distance from it equal to four inches. The extremities 
of the line ef will be found by erecting perpendiculars from e' 
and/', intersecting ef in the points / and / Connect the 
points e and a by an arc of a circle, described with a radius 
equal to five inches. Also connect the poin«ts / and (^ by a 
similar arc. efik is the elevation. 

To construct the isometric projection : assume a point a, and 
draw the projection of the pedestal, four inches by two inches 
by one-half inch, as in Prob. I., Plate XII. To find the point 
e, the projection of one point of the rectangle, the outline of 
the upper plane surface: produce h'e' until it intersects a'b' in 
/'. Also draw ^/perpendicular to ab. The distance from a' to 
/' is one inch and a half; lay off this distance from a, to / and 
draw the 30° line l,ni^. The distance from /' to e' is seven- 
eighths of an inch ; lay off this distance from /^ to in^ and draw 
the perpendicular m/,. The distance from / to ^ is three inches 
and a half; lay off this distance from m^ to e^ and construct the 
projection of the rectangle ^,f ^g,h^ one inch by one-quarter of 
an inch. The points cLpbpC^,d^,e^,f ^^g ^ and h, are the projections 
of the extremities of the curves. Intermediate points may be 
found as follows: suppose a plane passed through the solid, 
parallel to the base of the pedestal. This plane will cut from 
it a rectangle, the elevation of which will be a line up drawn 
parallel to ab, and the plan a rectangle n p'qr\ the sides 
n' p' and q'r' being parallel to a'b\ and the sides n'r' a.ndp'q' 
parallel to a'd'. The length n'p' is equal to np, and the width 
is determined by dropping perpendiculars from the points n 
and/, limiting them between the lines e'a' ,f'b' yg'c' and Ji d' . 



52 

Find the projection of this rectangle n^p^q^r,, by a process 
similar to that employed for finding the projection of the upper 
one, the construction of each being fully illustrated. The four 
angular points are points in the projections of the curves, any 
number of which may be found by similar constructions, which 
sbo M be taken near together in order that the curves may be 
accurately traced. 

PLATE XV. 

Problem. To dmw the isometric projection of a Pillow Block. 

Let the dimensions be those which are given in Plate VII. 
The construction will be made as follows : draw the 
projection of the pedestal twelve inches by three inches 
by one inch, as in Prob I., Plate XII. Bisect ac in 
l\ and erect the perpendicular bd equal to four inches. 
Through d draw the 30° line ef, and lay off de and dfy 
each equal to two inches and one-eighth. Draw the perpen- 
diculars eg and/>^. Lay off di and dj, each equal to one inch 
and a half, and draw the perpendiculars ik and//, each equal to 
one inch and five-eighths. Lay off bin equal to one inch and 
a half, and through m draw the 30° line no one inch and a 
quarter long, that is, lay off mn and mo, each equal to five- 
eighths of an inch. Join hi and lo, and from each of the 
points /j;*,/,^,/> and g draw 30° lines, each three inches long. 
From j^ draw the perpendicular j^l, intersecting // in /, ; 
and from l, draw a parallel to lo as far as it is visible. 
Bisect jf] at the point b, and draw the 30° line fiq. Make 
btf.ibq and fit each equal to one inch by the isovietric scale. 
Draw ,bs one inch long by the ordinary scale, parallel to the 
ground line. Draw the perpendicular fis^ and from s draw the 
30° line ss^ intersecting fis, in s^. bs is the semi-major axis, 
and bs^ the semi-minor axis of an ellipse, one-half of which 
will be the projection of the semi-circle tn'n'p'^ and this 
curve may be drawn as in Fig. I., Plate XL, to which the 
student's attention is called. The major axis is equal -to 
the diameter of the circle of which the ellipse is the projection, 
and its length in this particular case is laid off on the Hne 



53 

corresponding to ec drawn parallel to the ground line. From 
s draw a perpendiclar as far as it is visible. Similarly con- 
struct the semi-ellipse v^u^uv and the visibLe part of the 
curve WfWx. Draw the perpendiculars tt^Wf and vx. By 
similar constructions we may find the projections of the 
holes rq and r'q' , as follows : draw the centre line yz. Lay 
o^yOf a.ndyo^, equal respectively to one inch and a quarter, 
and one inch and seven-eighths. The point o, is the centre of 
a semi-ellipse, which will be drawn in a manner similar to the 
construction for the curve v^UiUv ; and the point 0^, is the 
centre of a semi-ellipse, which will be drawn in a manner sim- 
ilar to the construction for the curve tiS,qst. The major axis 
in each case will be equal to seven-eighths of an inch by the 
ordinary scale ; and seven-sixteenths of an inch by the isometric 
scale will be laid off from the centre on each of the 30° lines 
to determine other points of the curves. The vertex of the 
minor axis will be found as before illustrated. These semi- 
ellipses are connected by 30° lines which are tangent to the 
curves. Similarly the projection of the outline of the hole at 
the other end is found. 

Ex. VI. To draw the elevation of a solid whose plan is given. 

Let the plan be a rectangle a'a"h"h' , a copy of the 
plan, Prob. IV., Plate II., the lines a'a" and h'h" form- 
ing angles of 60°, and the lines a'h' and a"h" forming 
angles of 30° with the ground line. Draw all the lines, both 
visible and invisible, and find the elevation, it being understood 
that the distances from the horizontal edges to the horizontal 
plane are equal respectively to the distances from the points 
ciyCye^fy etc., to the ground line. 

Ex. VII. Draw the isometric projection of a pedestal ; 
dimensions given, Plate VIII., No. I. 

Ex. VIII. Draw the isometric projection of a crank ; 
dimensions given, Plate IX., No. I. 

Ex. IX. Draw the isometric projection of a crank shaft; 
dimensions given, Plate X., No. I. 



54 

Remark. In these exercises the student will have an 
opportunity of drawing the isometric projection of a circle in 
different positio;is. The major axis of the ellipse will be equal 
to the diameter of the circle of which it is the projection, and 
its direction will be the same as in one or other of the three 
cases, viz.: ac, ce or eg,, Fig. I., Plate XL, to which attention is 
especially directed. 



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